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SCHAUM'S OUTLINE OF 1
THEORY AND PROBLEMS
OF
THEORETICAL MECHANICS
with an introduction to
Lagrange's Equations
and Hamiltonian Theory
(i
BY
MURRAY R. SPIEGEL, Ph.D.
Professor of Mathematics
Rensselaer Polytechnic Institute
SCHAUM'S OUTLINE SERIES
McGRAWHILL BOOK COMPANY
New York, St. Louis, San Francisco, Toronto, Sydney
Copyright © 1967 by McGrawHill, Inc. All rights reserved. Printed in the
United States of America. No part of this publication may be reproduced,
stored in a retrieval system, or transmitted, in any form or by any means,
electronic, mechanical, photocopying, recording, or otherwise, without the prior
written permission of the publisher.
60232
8 9 10 11 12 13 14 15 SH SH 7 5
Preface
In the 17th century, Sir Isaac Newton formulated his now famous laws of mechanics.
These remarkably simple laws served to describe and predict the motions of observable
objects in the universe, including those of the planets of our solar system.
Early in the 20th century it was discovered that various theoretical conclusions de
rived from Newton's laws were not in accord with certain conclusions deduced from theories
of electromagnetism and atomic phenomena which were equally well founded experimentally.
These discrepancies led to Einstein's relativistic mechanics which revolutionized the con
cepts of space and time, and to quantum mechanics. For objects which move with speeds
much less than that of light and which have dimensions large compared with those of atoms
and molecules Newtonian mechanics, also called classical mechanics, is nevertheless quite
satisfactory. For this reason it has maintained its fundamental importance in science and
engineering.
It is the purpose of this book to present an account of Newtonian mechanics and its
applications. The book is designed for use either as a supplement to all current standard
textbooks or as a textbook for a formal course in mechanics. It should also prove useful to
students taking courses in physics, engineering, mathematics, astronomy, celestial me
chanics, aerodynamics and in general any field which needs in its formulation the basic
principles of mechanics.
Each chapter begins with a clear statement of pertinent definitions, principles and
theorems together with illustrative and other descriptive material. This is followed by
graded sets of solved and supplementary problems. The solved problems serve to illustrate
and amplify the theory, bring into sharp focus those fine points without which the student
continually feels himself on unsafe ground, and provide the repetition of basic principles
so vital to effective learning. Numerous proofs of theorems and derivations of basic re
sults are included in the solved problems. The large number of supplementary problems
with answers serve as a complete review of the material of each chapter.
Topics covered include the dynamics and statics of a particle, systems of particles and
rigid bodies. Vector methods, which lend themselves so readily to concise notation and to
geometric and physical interpretations, are introduced early and used throughout the book.
An account of vectors is provided in the first chapter and may either be studied at the be
ginning or referred to as the need arises. Added features are the chapters on Lagrange's
equations and Hamiltonian theory which provide other equivalent formulations of
Newtonian mechanics and which are of great practical and theoretical value.
Considerably more material has been included here than can be covered in most courses.
This has been done to make the book more flexible, to provide a more useful book of
reference and to stimulate further interest in the topics.
I wish to take this opportunity to thank the staff of the Schaum Publishing Company
for their splendid cooperation.
M. R. Spiegel
Rensselaer Polytechnic Institute
February, 1967
CONTENTS
Page
Chapter 1 VECTORS, VELOCITY AND ACCELERATION
Mechanics, kinematics, dynamics and statics. Axiomatic foundations of me
chanics. Mathematical models. Space, time and matter. Scalars and vectors.
Vector algebra. Laws of vector algebra. Unit vectors. Rectangular unit vec
tors. Components of a vector. Dot or scalar product. Cross or vector product.
Triple products. Derivatives of vectors. Integrals of vectors. Velocity. Ac
celeration. Relative velocity and acceleration. Tangential and normal acceler
ation. Circular motion. Notation for time derivatives. Gradient, divergence
and curl. Line integrals. Independence of the path. Free, sliding and bound
vectors.
Chapter 2 NEWTON'S LAWS OF MOTION. WORK, ENERGY
AND MOMENTUM 33
Newton's laws. Definitions of force and mass. Units of force and mass.
Inertial frames of reference. Absolute motion. Work. Power. Kinetic energy.
Conservative force fields. Potential energy or potential. Conservation of
energy. Impulse. Torque and angular momentum. Conservation of momentum.
• Conservation of angular momentum. Nonconservative forces. Statics or equi
librium of a particle. Stability of equilibrium.
Chapter 6 MOTION IN A UNIFORM FIELD. FALLING BODIES
AND PROJECTILES
Uniform force fields. Uniformly accelerated motion. Weight and acceleration
due to gravity. Gravitational system of units. Assumption of a flat earth.
Freely falling bodies. Projectiles. Potential and potential energy in a uniform
force field. Motion in a resisting medium. Isolating the system. Constrained
motion. Friction. Statics in a uniform gravitational field.
62
Chapter 4 THE SIMPLE HARMONIC OSCILLATOR AND
THE SIMPLE PENDULUM 86
The simple harmonic oscillator. Amplitude, period and frequency of simple
harmonic motion. Energy of a simple harmonic oscillator. The damped har
monic oscillator. Overdamped, critically damped and underdamped motion.
Forced vibrations. Resonance. The simple pendulum. The two and three
dimensional harmonic oscillator.
Chapter 5 CENTRAL FORCES AND PLANETARY MOTION 116
Central forces. Some important properties of central force fields. Equations
of motion for a particle in a central field. Important equations deduced from
the equations of motion. Potential energy of a particle in a central field. Con
servation of energy. Determination of the orbit from the central force. Deter
mination of the central force from the orbit. Conic sections, ellipse, parabola
and hyperbola. Some definitions in astronomy. Kepler's laws of planetary
motion. Newton's universal law of gravitation. Attraction of spheres and
other objects. Motion in an inverse square field.
CONTENTS
Page
Chapter 6 MOVING COORDINATE SYSTEMS 144
Noninertial coordinate systems. Rotating coordinate systems. Derivative
operators. Velocity in a moving system. Acceleration in a moving system.
Coriolis and centripetal acceleration. Motion of a particle relative to the earth.
Coriolis and centripetal force. Moving coordinate systems in general. The
Foucault pendulum.
Chapter 7 SYSTEMS OF PARTICLES 165
Discrete and continuous systems. Density. Rigid and elastic bodies. Degrees
of freedom. Center of mass. Center of gravity. Momentum of a system of
particles. Motion of the center of mass. Conservation of momentum. Angular
momentum of a system of particles. Total external torque acting on a system.
Relation between angular momentum and total external torque. Conservation
of angular momentum. Kinetic energy of a system of particles. Work. Po
tential energy. Conservation of energy. Motion relative to the center of mass.
Impulse. Constraints. Holonomic and nonholonomic constraints. Virtual dis
placements. Statics of a system of particles. Principle of virtual work. Equi
librium in conservative fields. Stability of equilibrium. D'Alembert's principle.
Chapter 8 APPLICATIONS TO VIBRATING SYSTEMS,
ROCKETS AND COLLISIONS 194
Vibrating systems of particles. Problems involving changing mass. Rockets.
Collisions of particles. Continuous systems of particles. The vibrating string.
Boundaryvalue problems. Fourier series. Odd and even functions. Con
vergence of Fourier series.
Chapter 9 PLANE MOTION OF RIGID BODIES 224
Rigid bodies. Translations and rotations. Euler's theorem. Instantaneous
axis of rotation. Degrees of freedom. General motion of a rigid body. Chasle's
theorem. Plane motion of a rigid body. Moment of inertia. Radius of gyra
tion. Theorems on moments of inertia. Parallel axis theorem. Perpendicular
axes theorem. Special moments of inertia. Couples. Kinetic energy and
angular momentum about a fixed axis. Motion of a rigid body about a fixed
axis. Principle of angular momentum. Principle of conservation of energy.
Work and power. Impulse. Conservation of angular momentum. The com
pound pendulum. General plane motion of a rigid body. Instantaneous center.
Space and body centrodes. Statics of a rigid body. Principle of virtual work
and D'Alembert's principle. Principle of minimum potential energy. Stability.
Chapter 10 SPACE MOTION OF RIGID BODIES 253
General motion of rigid bodies in space. Degrees of freedom. Pure rotation
of rigid bodies. Velocity and angular velocity of a rigid body with one point
fixed Angular momentum. Moments of inertia. Products of inertia. Moment
of inertia matrix or tensor. Kinetic energy of rotation. Principal axes of
inertia. Angular momentum and kinetic energy about the principal axes. The
ellipsoid of inertia. Euler's equations of motion. Force free motion. The in
variable line and plane. Poinsot's construction. Polhode. Herpolhode. Space
and body cones. Symmetric rigid bodies. Rotation of the earth. The Euler
angles. Angular velocity and kinetic energy in terms of Euler angles. Motion
of a spinning top. Gyroscopes.
CONTENTS
Page
Chapter // LAGRANGE'S EQUATIONS 282
General methods of mechanics. Generalized coordinates. Notation. Trans
formation equations. Classification of mechanical systems. Scleronomic and
rheonomic systems. Holonomic and nonholonomic systems. Conservative and
nonconservative systems. Kinetic energy. Generalized velocities. Gener
alized forces. Lagrange's equations. Generalized momenta. Lagrange's equa
tions for nonholonomic systems. Lagrange's equations with impulsive forces.
Chapter 12 HAMILTONIAN THEORY 311
Hamiltonian methods. The Hamiltonian. Hamilton's equations. The Hamil
tonian for conservative systems. Ignorable or cyclic coordinates. Phase space.
Liouville's theorem. The calculus of variations. Hamilton's principle. Can
onical or contact transformations. Condition that a transformation be canoni
cal. Generating functions. The HamiltonJacobi equation. Solution of the
HamiltonJacobi equation. Case where Hamiltonian is independent of time.
Phase integrals. Action and angle variables.
APPENDIX A
UNITS AND DIMENSIONS 339
APPENDIX B
ASTRONOMICAL DATA 342
appendix c SOLUTIONS OF SPECIAL DIFFERENTIAL EQUATIONS .... 344
APPENDIX D INDEX OF SPECIAL SYMBOLS AND NOTATIONS 356
INDEX 361
and ACCELERATION
MECHANICS, KINEMATICS, DYNAMICS AND STATICS
Mechanics is a branch of physics concerned with motion or change in position of
physical objects. It is sometimes further subdivided into:
1. Kinematics, which is concerned with the geometry of the motion,
2. Dynamics, which is concerned with the physical causes of the motion,
3. Statics, which is concerned with conditions under which no motion is apparent.
AXIOMATIC FOUNDATIONS OF MECHANICS
An axiomatic development of mechanics, as for any science, should contain the following
basic ingredients:
1. Undefined terms or concepts. This is clearly necessary since ultimately any
definition must be based on something which remains undefined.
2. Unproved assertions. These are fundamental statements, usually in mathematical
form, which it is hoped will lead to valid descriptions of phenomena under study.
In general these statements, called axioms or postulates, are based on experimental
observations or abstracted from them. In such case they are often called laws.
3. Defined terms or concepts. These definitions are given by using the undefined
terms or concepts.
4. Proved assertions. These are often called theorems and are proved from the
definitions and axioms.
An example of the "axiomatic way of thinking" is provided by Euclidean geometry in
which point and line are undefined concepts.
MATHEMATICAL MODELS
A mathematical description of physical phenomena is often simplified by replacing
actual physical objects by suitable mathematical models. For example in describing the
rotation of the earth about the sun we can for many practical purposes treat the earth
and sun as points.
SPACE, TIME AND MATTER
From everyday experience, we all have some idea as to the meaning of each of the
following terms or concepts. However, we would certainly find it difficult to formulate
completely satisfactory definitions. We take them as undefined concepts.
2 VECTORS, VELOCITY AND ACCELERATION [CHAP. 1
1. Space. This is closely related to the concepts of point, position, direction and
displacement. Measurement in space involves the concepts of length or distance,
with which we assume familiarity. Units of length are feet, meters, miles, etc.
In this book we assume that space is Euclidean, i.e. the space of Euclid's geometry.
2. Time. This concept is derived from our experience of having one event taking
place after, before or simultaneous with another event. Measurement of time is
achieved, for example, by use of clocks. Units of time are seconds, hours, years, etc.
3. Matter. Physical objects are composed of "small bits of matter" such as atoms
and molecules. From this we arrive at the concept of a material object called a
particle which can be considered as occupying a point in space and perhaps moving
as time goes by. A measure of the "quantity of matter" associated with a particle
is called its mass. Units of mass are grams, kilograms, etc. Unless otherwise
stated we shall assume that the mass of a particle does not change with time.
Length, mass and time are often called dimensions from which other physical quantities
are constructed. For a discussion of units and dimensions see Appendix A, Page 339.
SCALARS AND VECTORS
Various quantities of physics, such as length, mass and time, require for their specifica
tion a single real number (apart from units of measurement which are decided upon in
advance). Such quantities are called scalar s and the real number is called the magnitude
of the quantity. A scalar is represented analytically by a letter such as t, m, etc.
Other quantities of physics, such as displacement, require for their specification a
direction as well as magnitude. Such quantities are called vectors. A vector is repre
sented analytically by a bold faced letter such as A in Fig. 11. Geometrically it is
represented by an arrow PQ where P is called the initial point and Q is called the terminal
point. The magnitude or length of the vector is then denoted by A or A.
Fig. 11 Fig. 12 Fig. 13
VECTOR ALGEBRA
The operations of addition, subtraction and multiplication familiar in the algebra of
real numbers are with suitable definition capable of extension to an algebra of vectors.
The following definitions are fundamental.
1. Two vectors A and B are equal if they have the same magnitude and direction
regardless of their initial points. Thus A = B in Fig. 12 above.
2. A vector having direction opposite to that of vector A but with the same length is
denoted by A as in Fig. 13 above.
3. The sum or resultant of vectors A and B of Fig. l4(a) below is a vector C formed
by placing the initial point of B on the terminal point of A and joining the initial
point of A to the terminal point of B [see Fig. l4(b) below]. We write C = A + B.
This definition is equivalent to the parallelogram law for vector addition as indicated
in Fig. l4(c) below.
CHAP. 11
VECTORS, VELOCITY AND ACCELERATION
Fig. 14
Extensions to sums of more than two vectors are immediate. For example,
Fig. 15 below shows how to obtain the sum or resultant E of the vectors A, B, C
and D.
5.
rrfB+c+D
Fig. 15
The difference of vectors A and B, represented by A — B, is that vector C which
when added to B gives A. Equivalently, AB may be denned as A + (B). If
A = B, then A — B is defined as the null or zero vector represented by 0. This has
a magnitude of zero but its direction is not defined.
The product of a vector A by a scalar p is a vector pA or Ap with magnitude
\p\ times the magnitude of A and direction the same as or opposite to that of A
according as p is positive or negative. If p = 0, pA = 0, the null vector.
LAWS OF VECTOR ALGEBRA
If A, B and C are vectors, and p and q are scalars, then
1. A + B = B + A
2. A + (B + C) = (A + B) + C
3. p(qA) = (pq)A = q(pA)
4. (p + q)A = pA + qA
5. p{A + B) = pA + pB
Commutative Law for Addition
Associative Law for Addition
Associative Law for Multiplication
Distributive Law
Distributive Law
Note that in these laws only multiplication of a vector by one or more scalars is
defined. On pages 4 and 5 we define products of vectors.
UNIT VECTORS
Vectors having unit length are called unit vectors. If A is a vector with length A > 0,
then A/A = a is a unit vector having the same direction as A and A = A a.
RECTANGULAR UNIT VECTORS
The rectangular unit vectors i, j and k are mutually perpendicular unit vectors having
directions of the positive x, y and z axes respectively of a rectangular coordinate system
VECTORS, VELOCITY AND ACCELERATION
[CHAP. 1
[see Fig. 16]. We use righthanded rectangular coordinate
systems unless otherwise specified. Such systems derive
their name from the fact that a right threaded screw ro
tated through 90° from Ox to Oy will advance in the posi
tive z direction. In general three vectors A, B and C which
have coincident initial points and are not coplanar are said
to form a righthanded system or dextral system if a right
threaded screw rotated through an angle less than 180° from
A to B will advance in the direction C [see Fig. 17 below].
\/o.
w v
Fig. 16
(4i,A2,As)
Fig. 17
Fig. 18
COMPONENTS OF A VECTOR
Any vector A in 3 dimensions can be represented with initial point at the origin O of
a rectangular coordinate system [see Fig. 18 above]. Let (Ai,A 2 ,A 3 ) be the rectangular
coordinates of the terminal point of vector A with initial point at 0. The vectors Aii,
A 2 j and A 3 k are called the rectangular component vectors, or simply component vectors,
of A in the x, y and z directions respectively. A lf A 2 and A 3 are called the rectangular
components, or simply components, of A in the x, y and z directions respectively.
The sum or resultant of Aii, A 2 j and Ask is the vector A, so that we can write
A = Aii + A 2 j + A 3 k
(1)
The magnitude of A is
A = A = v^f+AfTAf
In particular, the position vector or radius vector r from O to the point {x,y,z) is
written . , . , .
r = xi + yj + zk
(2)
is
(3)
and has magnitude r = r = y/x 2 + y 2 + z 2 .
DOT OR SCALAR PRODUCT
The dot or scalar product of two vectors A and B, denoted by AB (read A dot B)
is defined as the product of the magnitudes of A and B and the cosine of the angle
between them. In symbols,
A«B = AS cos 0, 0^0^77 (4)
Note that A*B is a scalar and not a vector.
CHAP. 11
VECTORS, VELOCITY AND ACCELERATION
The following laws are valid:
1. A • B = B • A Commutative Law for Dot Products
2. A«(B + C) = A»B + A«C Distributive Law
3. p{A • B) = (pA) • B = A • (pB) = (A • B)p, where p is a scalar.
4. i*i = j*j = k*k = l, i*j = j*k = k*i =
5. If A = Aii + Aaj + Aak and B = Bd + B 2 j + Bak, then
A • B = AiBi + A2B2 + A3B3
A A = A 2 = A\ + A\ + A\
BB = B 2 = B\ + B\ + B%
6. If AB = and A and B are not null vectors, then A and B are perpendicular.
CROSS OR VECTOR PRODUCT
The cross or vector product of A and B is a vector C = A x B (read A cross B). The
magnitude of A x B is denned as the product of the magnitudes of A and B and the sine
of the angle between them. The direction of the vector C = A x B is perpendicular to the
plane of A and B and such that A, B and C form a righthanded system. In symbols,
A X B = AB sin 6 u, ^ 6 ^ tt (5)
where u is a unit vector indicating the direction of A x B. If A = B or if A is parallel
to B, then sin 6 = and we define A x B = 0.
The following laws are valid:
1.
2.
3.
4.
5.
A x B = — B x A (Commutative Law for Cross Products Fails)
Ax(B + C) = AxB + AxC Distributive Law
p{A x B) = {pA) x B = A x (pB)  (A x B)p, where p is a scalar.
ixi = jxj = kxk = 0, ixj = k, j x k = i, kxi = j
If A = Aii + Aaj + Aak and B = Bd + B 2 j +
AxB =
1
At
Bx
1
A 2
B 2
then
k
A 3
B 3
6.
7.
A x B = the area of a parallelogram with sides A and B.
If A x B = and A and B are not null vectors, then A and B are parallel.
TRIPLE PRODUCTS
The scalar triple product is defined as
At
A 2
As
• (B x C) =
Bx
B 2
B z
Ct
C 2
Cz
(')
where A = Aii + A 2 j + A3k, B = B\\ + B 2 j + Bak, C = Cd + C 2 j + Cak. It represents the
volume of a parallelepiped having A, B, C as edges, or the negative of this volume according
as A, B, C do or do not form a right handed system. We have A • (B x C) = B • (C x A) =
C(AXB).
The vector triple product is defined as
AX(BXC) = (AC)B(AB)C {7)
Since (A x B) x C = (A • C)B  (B • C) A, it is clear that A x (B x C) ¥• (A x B) x C.
6 VECTORS, VELOCITY AND ACCELERATION [CHAP. 1
DERIVATIVES OF VECTORS
If to each value assumed by a scalar variable u there corresponds a vector A(u), or
briefly A, then A(u) is called a (vector) function of u. The derivative of A(u) is defined as
dU Au~0 AW V '
provided this limit exists. If A(u) = Ai(u)i + Ai{u)i + A 3 (w)k, then
dA _ dAi . dAi. c?A 3 . ,q*
du ~~ du du du
Similarly we can define higher derivatives. For example the second derivative of A(u)
if it exists is given by
&A _ ^Ai. &A2. d?As , /m
du 2 ~ du* 1 + du 2 3 + « K I 1 "*
Example. If A = (2m 2 — 3tt)i + 5 cos u j — 3 sin u k, then
^ = (4m  3)i  5 sin m j — 3 cos u k, jg = 4i  5 cos u j + 3 sin u k
The usual rules of differentiation familiar in the calculus can be extended to vectors,
although order of factors in products may be important. For example if ${u) is a scalar
function while A{u) and B(u) are vector functions, then
t(**) = *•£ + £•» <">
A(AXB) = Axf + fxB (IS)
INTEGRALS OF VECTORS
Let A{u) = Ai(u)i + A 2 (u)j + A s (u)k be a vector function of u. We define the indefinite
integral of A(u) as
( A(u)du = i f A x {u)du + j J A 2 (w)^ + kj A 3 (u)du (U)
If there exists a vector function B(u) such that A(w) = ^{B(w)}, then
J*A(M)dM = r^{B(tt)}<fa = B(m) + c (*5)
where c is an arbitrary constant vector independent of u. The definite integral between
limits u = a and u = /? is in such case, as in elementary calculus, given by
C A(u)du = f'fo{B{u))du = B(u) + c ' = B(/3)  B(«) (iff)
The definite integral can also be defined as a limit of a sum analogous to that of elementary
calculus.
VELOCITY
Suppose that a particle moves along a path or curve C [Fig. 19 below]. Let the position
vector of point P at time t be r = r(t) while the position vector of point Q at time t + At is
CHAP. 11
VECTORS, VELOCITY AND ACCELERATION
r + Ar = r(t + At). Then the velocity (also called the instantaneous velocity) of the particle
at P is given by
dr .. Ar
v = — = hm — 
at At*o At
At*o AC
and is a vector tangent to C at P.
(17)
If r = r(t) = x(t)i + y(t)j + z{t)k = xi + yj + zk,
we can write
dt dx. dy . dz . .
v = dt = dt 1 + ^ 3 + Tt k ( 18)
The magnitude of the velocity is called the speed and is given by
Fig. 19
dr
dt
 V(t y+ (f r ♦ (t ) 2 =
V = v =
where s is the arc length along C measured from some initial point to P.
dt
(19)
ACCELERATION
If v = dr/dt is the velocity of the particle, we define the acceleration (also called the
instantaneous acceleration) of the particle at P as
a = dv = Hm v(t + At)v(t)
dt At»0 At
In terms of r = xi + yj + zk the acceleration is
_ d^r _ d?x. d?y.
* ~ dt* ~ dt 21 + dt 2 ~
and its magnitude is
i + 5^J + ^5k
dV
dt 2 '
a =
= V(S)' + (SWS
(21)
(22)
RELATIVE VELOCITY AND ACCELERATION
If two particles Pi and Pz are moving with respective velocities vi and V2 and accelera
tions ai and a2, the vectors
vpg/Pj = v 2  vi and ap 2 /Pj = a2  ai (23
are respectively called the relative velocity and relative acceleration of P2 with respect to Pi
TANGENTIAL AND
NORMAL ACCELERATION
Suppose that particle P with position vec
tor r = r(t) moves along curve C [Fig. 110].
We can consider a rectangular coordinate
system moving with the particle and defined
by the unit tangent vector T, the unit princi
pal normal N and the unit binormal B to
curve C where
Fig. 110
8
VECTORS, VELOCITY AND ACCELERATION
[CHAP. 1
T =
B = TxN
(U)
* N = R M
ds' * K ds'
s being the arc length from some initial point to P and R the radius of curvature of C at P.
The reciprocal of the radius of curvature is called the curvature and is given by k= 1/R.
We can show [see Problem 1.35, page 20] that the acceleration along C is given by
dt L + R* 1
(25)
The first and second terms on the right are called the tangential acceleration and normal
or centripetal acceleration respectively.
CIRCULAR MOTION
Suppose particle P moves on a circle C of radius
R. If s is the arc length measured along C from
A to P and 9 is the corresponding angle subtended
at the center O, then s = Rd. Thus the magnitudes
of the tangential velocity and acceleration are given
respectively by
ft = fi » (« e >
v
at
and
dv _ d?s _ p d 2
dt dt 2 K dt 2
= Ra
(27)
Fig. 111
We call © = dd/dt and a  d 2 9/dt 2 the angular speed and angular acceleration respectively.
The normal acceleration as seen from (25) is given by v 2 /R = <a 2 R.
NOTATION FOR TIME DERIVATIVES
We shall sometimes find it convenient to use dots placed over a symbol to denote
derivatives with respect to time t, one dot for a first derivative, two dots for a second
derivative, etc. Thus for example r = dr/dt, r = d 2 r/dt 2 , v = dvldt, etc.
GRADIENT, DIVERGENCE AND CURL
If to each point (x, y, z) of a rectangular coordinate system there corresponds a vector A,
we say that A = A(x, y, z) is a vector function of x, y, z. We also call A(x, y, z) a vector
field. Similarly we call the (scalar) function <f>(x, y, z) a scalar field.
It is convenient to consider a vector differential operator called del given by
Using this we define the following important quantities.
= ( 1 a^ + J ^ + k ^
1. Gradient V4> = (i^ + J^7 + k ^j<^  * dx
This is a vector called the gradient of <£ and is also written grad </>.
+ 3 dy + K dz
(28)
(29)
2. Divergence
V'A =
l Jx + z dy dz
Mi + Mi + M§
a« a?/ dz
(Ad + A 2 j + AOt)
This is a scalar called the divergence of A and is also written div A.
(SO)
CHAP. 1]
VECTORS, VELOCITY AND ACCELERATION
3. Curl
V x A =
i 5S + ^ + k £) x < A ' i + A2i + A * )
i
j
k
d
d
d
dX
dy
dz
At
A 2
As
(31)
fdAs
\dy
BA
dz
„ l + ( g
dA s \. , fdA 2
dX
j + [iFf ]k
This is a vector called the curl of A and is also written curl A.
Two important identities are
divcurlA = V(V'XA) =
curl grad <£ = V x (V4>) =
(32)
LINE INTEGRALS
Let r(t) = x(t)i + y(t)j + z(t)k, where r(t) is the position vector of (x,y,z), define a
curve C joining points Pi and Pi corresponding to t = U and t = t 2 respectively. Let
A = A(x, y, z) = Ad + A 2 j + Ask be a vector function of position (vector field). The
integral of the tangential component of A along C from Pi to P%, written as
JA'dr = ( A«dr = I Aidx + A 2 dy + A 3 dz
(84)
is an example of a line integral.
If C is a closed curve (which we shall suppose is a simple closed curve, i.e. a curve
which does not intersect itself anywhere) then the integral is often denoted by
£ Adr = <k Aidx + A 2 dy + A s dz
(35)
In general, a line integral has a value which depends on the path. For methods of
evaluation see Problems 1.39 and 1.40.
INDEPENDENCE OF THE PATH
The line integral (34) will be independent of the path joining Pi and P 2 if and only if
A = v<£, or equivalently V x A = 0. In such case its value is given by
J»p 2 x»p 2
Adr = I d<f> = <j>(P 2 )  4>(Pi) = <f>(x 2 ,y 2 ,z 2 )  <f>(x u yi,Zx)
Pt J Pi
(36)
assuming that the coordinates of Pi and P 2 are (x v y v z t ) and (x 2 , y 2 , z 2 ) respectively while
<f>(x,y,z) has continuous partial derivatives. The integral (35) in this case is zero.
FREE, SLIDING AND BOUND VECTORS
Up to now we have dealt with vectors which are specified by magnitude and direction
only. Such vectors are called free vectors. Any two free vectors are equal as long as
they have the same magnitude and direction [see Fig. l12(a) below].
10
VECTORS, VELOCITY AND ACCELERATION
[CHAP. 1
(a) Equal free vectors
(b) Equal sliding vectors
Fig. 112
(c) Bound vector
Sometimes in practice the particular line of action of a vector is important. In such
case two vectors are equal if and only if they have the same magnitude, direction and line
of action. Such vectors are often called sliding vectors [see Fig. 112(6)].
Sometimes it is important to specify the point of action of a vector. Such a vector
[see Fig. l12(c)] is called a bound vector. In this case two vectors will be equal if and
only if they are identical.
Most cases with which we shall deal involve free vectors. Cases where sliding vectors
or bound vectors need to be employed will in general be clear from the context.
Solved Problems
VECTOR ALGEBRA
1.1. Show that addition of vectors is commutative, i.e. A + B = B + A. See Fig. 113
below.
OP + PQ . = OQ or A + B = C
and OR + RQ = OQ or B + A = C
Then A + B = B + A.
1.2. Show that the addition of vectors is associative, i.e. A + (B + C) = (A 4 B) + C. See
Fig. 114 above.
OP + PQ = OQ = (A + B) and PQ + QR = PR = (B + C)
Since OP + PR = OR = D, i.e. A + (B + C) = D
OQ + QR = OR = D, i.e. ( A + B) + C = D
we have A + (B + C) = ( A + B) + C.
Extensions of the results of Problems 1.1 and 1.2 show that the order of addition of any
number of vectors is immaterial.
CHAP. 1]
VECTORS, VELOCITY AND ACCELERATION
11
1.3. Given vectors A, B and C [Fig. l15(a)] construct (a) A  B + 2C, (b) 3C  £(2A  B).
(a)
B
Pig. 115
1.4. Prove that the magnitude A of the vector A =
Ad + Aaj + Ask is A = y/A\ + A\ + A\. See
Fig. 116.
By the Pythagorean theorem,
(OP)* = (OQ)* + (QP) 2
where OP denotes the magnitude of vector OP, etc.
Similarly, (OQ)* = (OR)* + (RQ)*.
Then (OP) 2 = (OR)* + (RQ)* + (QP)* or
A* = A\ + A\ + A\, i.e. A = VA* + A* + A*
(2AB)
1.5. Determine the vector having the initial point
P(xi,yi,zi) and the terminal point Q(* 2 , 3/2, z 2 )
and find its magnitude. See Fig. 117.
The position vector of P is r t = x x \ + 2/ x j + Zjk.
The position vector of Q is r 2 = x 2 i + 2/ 2 J + «2 k 
r x + PQ = r 2 or
PQ = 1*2 — r i = (^2* + Vzi + *2 k ) — (M + #ii + *i k )
= («2  *i)i + (Va  2/i)J + («2  «i)k
Magnitude of PQ = PQ
= V(*2  *l) 2 + (^2 ~ Vl) 2 + (*2 ~ *l) 2
Note that this is the distance between points P and Q.
Fig. 117
12
VECTORS, VELOCITY AND ACCELERATION
[CHAP. 1
1.6. Find (a) graphically and (b) analytically the sum or resultant of the following
displacements:
A, 10 ft northwest; B, 20 ft 30° north of east; C, 35 ft due south. See Fig. 118.
Graphically.
At the terminal point of A place the initial point
of B. At the terminal point of B place the initial
point of C.
The resultant D is formed by joining the initial
point of A to the terminal point of C, i.e. D =
A + B + C.
The resultant is measured to have magnitude of
4.1 units = 20.5 ft and direction 60° south of east.
Analytically.
From Fig. 118, if i and j are unit vectors in the
E and N directions, we have
A =  10 cos 45° i + 10 sin 45° j
B = 20 cos 30° i + 20 sin 30° j
C = — 35j
Then the resultant is
D = A + B + C = (10 cos 45° + 20cos30°)i + (10 sin 45° + 20 sin 30°  35)j
= (5V2 + 10VS )i + (5 V^ + 10  35)j = 10.25i  17.93J
Unit = 5 ft
Thus the magnitude of D is V(10.25) 2 + (17.93) 2 = 20.65 ft and the direction is
tani 17.93/10.25 = tan" 1 1.749 = 60°45' south of east
Note that although the graphical and analytical results agree fairly well, the analytical result is
of course more accurate.
THE DOT OR SCALAR PRODUCT
1.7. Prove that the projection of A on B is equal to A • b,
where b is a unit vector in the direction of B.
Through the initial and terminal points of A pass
planes perpendicular to B at G and H respectively as in the
adjacent Fig. 119; then
Projection of A on B = GH = EF = A cos 6 = A • b
G H B
Fig. 119
1.8. Prove A«(B + C) = AB + AC.
Let a be a unit vector in the direction of A; then [see
Fig. 120]
Projection of (B + C) on A = projection of B on A
+ projection of C on A
(B + C)»a = B*a + C*a
Multiplying by A,
(B+C)'Aa = B»Aa + OAa
and (B + C) • A = B • A + C • A
Then by the commutative law for dot products,
A • (B + C) = A • B + A • C
and the distributive law is valid.
Fig. 120
CHAP. 1]
VECTORS, VELOCITY AND ACCELERATION
13
1.9. Evaluate each of the following.
(a) i«i = iicosO° = (1)(1)(1) = 1
(6) i»k = ikcos90° = (1)(1)(0) =
(e) kj = kjcos90° = (1)(1)(0) =
(d) j • (2i  8j + k) = 2j • i  3j • j + j • k = 03 + = 3
(e) (2i  j) • (3i + k) = 2i • (3i + k)  j • (3i + k) = 6i • i + 2i • k  3j • i  j • k
= 6 + 000 = 6
1.10. If A = Aii + Aaj+Agk and B = Bd + B 2 2 + B^k, prove that AB = Ai#i + A 2 JB 2 +
AsBa.
AB = (Aii + Aa + AMiBj + Bzi + Bak)
= A x \ • (BJ + Baj + B 3 k) + A 2 j • (£ x i + B 2 j + B 3 k) + A 3 k • (BJ + B 2 j + B 3 k)
= A x B x i • i + A^i • j + A X B Z \ • k + A 2 Btf • i + A 2 £ 2 j • j + A 2 £ 3 j • k
+ A^k • i + A 3 £ 2 k • j + A 3 £ 3 k • k
= A X B X + A 2 B % + A 3 fi 3
since i • i = j • j = k • k = 1 and all other dot products are zero.
1.11. If A = Aii + A 2 j + A 3 k, show that A = \/AA = y/A\ + Al + A\.
A«A = (A)(A) cos 0° = A 2 . Then A = VA • A.
Also, A»A = (Aji + A 2 j + A 3 k) • (A 1 i + A 2 j + A 3 k)
= (AJiAJ + (A 2 )(A 2 ) + (A S )(A S ) = A\ + A\ + Al
by Problem 1.10, taking B = A.
Then A = y/A* A = ^A\ + A 2 + A\ is the magnitude of A. Sometimes A • A is written A 2 .
1.12. Find the acute angle between the diagonals of a
quadrilateral having vertices at (0, 0, 0), (3, 2, 0),
(4,6,0), (1,3,0) [Fig. 121].
We have OA = 3i + 2j, OB = 4i + 6j, OC = i + 3j
from which
CA = OAOC = 2i — j
Then OB • CA = [OB CA cos $
i.e.
(4i + 6j) • (2i  j) = V(4) 2 + (6)2 V(2) 2 + (1) 2 cos e
from which cos 6 = 2/(v1S2 Vs ) = 1240 and = 82°53'.
5(4,6,0)
(0,0,0)
Fig. 121
THE CROSS OR VECTOR PRODUCT
1.13. Prove AxB = BxA.
Fig. 122
BXA = D
(&)
A X B = C has magnitude AB sin $ and direction such that A, B and C form a righthanded
system [Fig. l22(a) above].
14
VECTORS, VELOCITY AND ACCELERATION
[CHAP. 1
B X A = D has magnitude BA sin e and direction such that B, A and D form a righthanded
system [Pig. 122(6) above].
Then D has the same magnitude as C but is opposite in direction, i.e. C = — D or A X B = B X A.
The commutative law for cross products is not valid.
1.14. Prove that
Ax(B + C) = AXB + AXC
for the case where A is perpendicular
to B and also to C.
Since A is perpendicular to B, A X B is a
vector perpendicular to the plane of A and B
and having magnitude AB sin 90° = AB or
magnitude of AB. This is equivalent to mul
tiplying vector B by A and rotating the
resultant vector through 90° to the position
shown in Fig. 123.
Similarly, A X C is the vector obtained by
multiplying C by A and rotating the resultant
vector through 90° to the position shown.
In like manner, A X (B + C) is the vector
obtained by multiplying B + C by A and rotat
ing the resultant vector through 90° to the
position shown.
Since A X (B + C) is the diagonal of the
parallelogram with A X B and A X C as sides,
we have AX(BfC) = AXB + AXC.
Fig. 123
1.15. Prove that A x (B + C) = AxB + AxC
in the general case where A, B and C are
noncoplanar. See Fig. 124.
Resolve B into two component vectors, one
perpendicular to A and the other parallel to A,
and denote them by B^ and B (  respectively.
Then B = B ± + B„.
If 6 is the angle between A and B, then
B j_ = B sin e. Thus the magnitude of A X B j_ is
AB sin 9, the same as the magnitude of A X B.
Also, the direction of A X B ± is the same as the
direction of A X B. Hence A X B ± =AXB.
Similarly if C is resolved into two component
vectors C (  and C^, parallel and perpendicular
respectively to A, then A X C ± = A X C. Fig. 124
Also, since B + C = B ± + B n + C ± + C M = (Bj_ + C ± ) + (B, ( + C M ) it follows that
AX(B ± + C 1 ) = AX(B + C)
Now Bjl and Cj_ are vectors perpendicular to A and so by Problem 1.14,
AX(B 1 + C 1 ) = AXB i +AXC 1
Then A X (B + C) = AXB + AXC
and the distributive law holds. Multiplying by —1, using Problem 1.13, this becomes (B + C) X A =
B X A + C X A. Note that the order of factors in cross products is important. The usual laws of
algebra apply only if proper order is maintained.
i 3 k
1.16. If A = Ad + A 2 j + Ask and B = Bii + B 2 j + Bak, prove that A x B
Ai
A 2
B 2
As
Bs
CHAP. 1]
VECTORS, VELOCITY AND ACCELERATION
15
AXB = (Aii + Atf + AMxiBii + Btf + BJi.)
= Aii X (S x i + Ba + B s k) + Aa X (BJ + Brf + £ 3 k) + A 3 k X (BJ + # 2 j + B 3 k)
= A^i X i + A x B 2 i X j + A X B Z \ X k + AgBJ X i + A 2 5 2 j X J + ^2#3J X k
+ A 3 Bik X i + A 3 B 2 k X j + A 3 B 3 k X k
= (A 2 B 3  A 3 B 2 )i + (A s B t  A X B Z )\ + (A t B 2  A^k =
1.17. If A = 3ij + 2k and B = 2i + 3jk, findAxB.
j k
i
At
B,
j
k
A 2
^3
B 2
*3
AXB =
1 2
3 1
1
2
3
2
3
1
= 1
— J
+ k
3
1
2
1
2
3
= 5i + 7j + Ilk
1.18. Prove that the area of a parallelogram with
sides A and B is AxB.
Area of parallelogram = h B
=  A sin 6 B
= AXB
Note that the area of the triangle with sides A and
B is £ A X B .
B
Fig. 125
= £19i4j + 7k
1.19. Find the area of the triangle with vertices at P(2,3,5), Q(4,2,l), 72(3,6,4).
PQ = (4_ 2 )i + (23)j + (l5)k = 2ij6k
PR = (3  2)1 + (6  3)j + (4  5)k = i + 3 j  k
Area of triangle = £PQXPR = £ (2i j6k) X (i + 3jk)
i j k
= il 2 1 6
1 31
= iV(19) 2 + (4)2 + (7)2 = ■Jv'426
TRIPLE PRODUCTS
1.20. Show that A • (B x C) is in absolute value equal
to the volume of a parallelepiped with sides
A, B and C.
Let n be a unit normal to parallelogram /, having
the direction of B X C, and let h be the height of the
terminal point of A above the parallelogram /. Fig. 126
Volume of parallelepiped = (height fc)(area of parallelogram J)
= (A.n)(BXC)
= A»{BXCn} = A'(BXC)
If A, B and C do not form a righthanded system, A • n < and the volume =  A • (B X C)  .
1.21. (a) If A = Aii + A 2 j+Ask, B = Bd + B 2 j + Bsk, C = Cii + C 2 j + Ctk show that
Ai A 2 Az
A(BxC) = Bi B 2 B 3
C\ C 2 Ci
16
VECTORS, VELOCITY AND ACCELERATION
[CHAP. 1
(6) Give a geometric significance of the case where A • (B x C) = 0.
i j k
(a) A'(BXC) = A* B t B 2 B 3
Cj C 2 C 3
= (A i i + A 2 j + A 3 k) • [(B 2 C 3  £ 3 C 2 )i + {B 3 C X  B X C 3 )\ + (B X C 2  B 2 C t )k]
A x A 2 A3
= A 1 (B 2 C 3 B 3 C 2 )+A 2 (B 3 C 1 B 1 C 3 ) + A 3 {B X C 2  B 2 C t ) = B x B 2 B 3
C x C 2 c 3
(b) By Problem 1.20 if A • (B X C) = then A, B and C are coplanar, i.e. are in the same plane,
and conversely if A, B, C are coplanar then A • (B X C) = 0.
1.22. Find the volume of a parallelepiped with sides A = 3i — j, B = j + 2k, C = i + 5j + 4k.
310
By Problems 1.20 and 1.21, volume of parallelepiped = A»(BXC) =
12
15 4
= 1201 = 20.
1.23. If A = i + j, B = 2i3j + k, C = 4j3k, find (a)(AxB)xC, (b)Ax(BxC).
i j k
110
23 1
(a) A X B =
= i  j  5k. Then (A X B) X C =
i j k
1 1 5
43
= 23i + 3j + 4k.
(6) B X C =
i J k
23 1
43
= 5i + 6j + 8k. Then A X (B X C)
i
j
k
1
1
5
6
8
= 8i8j + k.
It follows that, in general, (A X B) X C ^ A X (B X C).
DERIVATIVES AND INTEGRALS OF VECTORS
1.24. If r = (i 8 + 2i)i3<? 2t j + 2sin5*k, find (a) ^, (b)
d 2 r
dt 2
at t = 0.
(a) ^ = ~(*s + 2t)i + ^(3e 2 *)j + ^(2 sin 5*)k = (3*2 + 2)i + 6e~2tj + 10 cos ht k
At * = 0, dr/dt = 2i + 6j + 10k.
(6) From (a), \dr/dt\ = V(2) 2 + (6) 2 + (10)2 = ^140 = 2\/35 at t = 0.
(c) W = Jtilt) = ^<( 3 * 2 + 2 ) i + 6e ~ 2t J + 10cos5«k} = 6<i12e2tj50sin5tk
At t =t 0, d?r/dtf = — 12j.
(d) From (c), Idtr/dP] =12 at t = 0.
1.25. Prove that f(AB) = A^ + ^B, where A and B are differentiate func
tions of u.
du
du tint,
CHAP. 1] VECTORS, VELOCITY AND ACCELERATION 17
„ L , , d ,. „, .. (A + AA) • (B + AB)  A'B
Method 1. j(A»B) = lim * .
du v am*o Am
A'AB + AA'B + AA'AB
= lim ■ t
Au»0 AM
/ AB AA AA \ . dB dA
= ?™o{ A '^ + ^ B+ ^' AB ) = A S£ + 1S' B
Method 2. Let A = A x i + A 2 j + A 3 k, B = Bji + B^ + £ 3 k. Then
£< A " B > = ^A + aa + aw
^ <Zi? 2 df? 3 \ , (dA x .dA 2 dA* \
A *dir + A *~dV + A » *r; + {~dV Bl + ~dV B2 + ~du~ Bs )
. dB , dA
dt* du
d 2
1.26. If <£(#, y, z) = z 2 2/s and A = Sx 2 yi + yz 2 \  xzk, find ^^ (<f>A) at the point (1, 2, 1).
$A = {x 2 yz){Sx 2 yi + yz 2 \  xzk) = 3x*y 2 zi + x 2 y 2 z 3 j  x 3 2/s 2 k
T(*A) = ^(SajVrf + aV*^ «■»«*) = Sx*y 2 i + 3x 2 y 2 z 2 j  2x3yzk
02 OZ
3 2 _, 3
(*A) = ^ (3a^i/ 2 i + Sx 2 y 2 z 2 j  2x 3 wzk) = 6as 4 ]/i + 6»; 2 i/3 2 j  2a; 3 *k
dydz
If « = 1, y = 2, z = 1, this becomes 12i  12j + 2k.
1.27. Evaluate f A(w)dw if A(u) = (3w 2  l)i + (2i*  3) j + {Qu 2  4tt)k.
The given integral equals
I {(3m 2  l)i + (2m  3)j + (6m 2  4M)k} du
2
u=l
= (m 3  M)i + (m 2  3m) j + (2m 3  2M 2 )k
= {(8  2)i + (4  6)j + (16  8)k>  {(1  Di + (1  3)j + (2  2)k}
= 6i + 8k
VELOCITY AND ACCELERATION
1.28. A particle moves along a curve whose parametric equations are x = 3e~ 2t , y = 4 sin 3t,
2 = 5 cos St where t is the time.
(a) Find its velocity and acceleration at any time.
(6) Find the magnitudes of the velocity and acceleration at £ = 0.
(a) The position vector r of the particle is
r = xi + yj + zk = 3e~ 2t i + 4 sin St j + 5 cos Bt k
Then the velocity is
v = dt/dt = 6e 2t i + 12 cos Zt j  15 sin St k
and the acceleration is
a = dv/dt = d 2 v/dt 2 = 12e 2t i  36 sin St j  45 cos 3* k
(6) At t = 0, v = dr/dt = 6i + 12j and a = d 2 r/dt 2 = 12i45k. Then
magnitude of velocity at t = is V(6) 2 + (12) 2 = 6^5
magnitude of acceleration at * = is V(12) 2 + (45) 2 = 3 V241
18
VECTORS, VELOCITY AND ACCELERATION
[CHAP. 1
1.29. A particle travels so that its acceleration is given by
a = 2e t i + 5 cos t j  3 sin t k
If the particle is located at (1,3,2) at time t = and is moving with a velocity
given by 4i  3j + 2k, find (a) the velocity and (b) the displacement of the particle
at any time t>0.
/ v dPr dv
(°) a = no = jT = 2e'i + 5 cos * j  3 sin * k
8 = W = ~dt = 2e_1 + 5cos *
Integrating, v = J (2e"'i + 5 cos t j  3 sin t k) dt
= —2e~H + 5 sin t j + 3 cos t k + c t
Since v = 4i3j + 2k at t = 0, we have
4i  3j + 2k = 2i + 3k + c t or c t = 6i  3j  k
Then v = 2«*i + 5 sin t j + 3 cos t k + 6i  3j  k
= (6  2e*)i + (5 sin t  3)j + (3 cos *  l)k
(6) Replacing v by dr/dt in (I) and integrating, we have
r = J [(6  2e*)i + (5 sin *  3)j + (3 cos t  l)k) dt
= (6t + 2«~*)i  (5 cos * + 8«)j + (3 sin t  t)k + c 2
Since the particle is located at (1,3,2) at t = 0, we have r = i — 3 j + 2k at t = 0, so that
i  3j + 2k = 2i  5j + c 2 or c 2 = i + 2j + 2k
Thus r = (6t + 2«~*l)i + (2  5 cos *  3«)j + (3 sin «  t + 2)k {2)
(1)
RELATIVE VELOCITY AND ACCELERATION
1.30. An airplane moves in a northwesterly
direction at 125 mi/hr relative to the
ground, due to the fact that there is a
westerly wind [i.e. from the west] of
50 mi/hr relative to the ground. Deter
mine (a) graphically and (b) analyti
cally how fast and in what direction
the plane would have traveled if there
were no wind.
(a) Graphically.
Let W = wind velocity
V a = velocity of plane
with wind
V(, = velocity of plane
without wind.
Then [see Pig. 127] V a = V b + W or
Fig. 127
v 6 = v a w = V„ + (W).
V b has magnitude 6.5 units = 163 mi/hr and direction 33° north of west.
(6) Analytically.
Letting i and j be unit vectors in directions E and N respectively, we see from Fig. 127
that
V =  125 cos 45° i + 125 sin 45° j and W = 50i
Then V„ = V a  W = ( 125 cos 45°  50)i + 125 sin 45° j = 138.391 + 88.39J.
Thus the magnitude of V b is V(138.39) 2 + (88.39) 2 = 164.2 mi/hr and the direction is
tani 88.39/138.39 = tan~i .6387 = 32° 34' north of west.
CHAP. 1] VECTORS, VELOCITY AND ACCELERATION 19
1.31. Two particles have position vectors given by ri = 2ti  t 2 j + (St 2  4£)k and
r 2 = (U 2  12* + 4)i + Fj  3tk. Find (a) the relative velocity and (b) the relative
acceleration of the second particle with respect to the first at the instant where t = 2.
(a) The velocities of the particles at t2 are respectively
Vl = r t = 2i2tj + (6t4)k
v 2 = f 2 = (lOt  12)i + St 2 }  3k
= 2i  4j + 8k
t = 2
= 8i + 12j  3k
t = 2
Relative velocity of particle 2 with respect to particle 1
= y 2  Vl = (8i + 12j  3k)  (2i  4j + 8k) = 6i + 16j  Ilk
(6) The. accelerations of the particles at t = 2 are respectively
a i — v i = *i = ~2j + 6k
a 2 = v 2 = r 2 = lOi + 6tj
= 2j + 6k
t = 2
= 10i + 12j
t = 2
Relative acceleration of particle 2 with respect to particle 1
= a 2  a! = (lOi + 12j)  (2j + 6k) = lOi + 14j  6k
TANGENTIAL AND NORMAL ACCELERATION
1.32. Given a space curve C with position vector
r = 3cos2ti'+ 3 sin 2t j + (8*4)k
(a) Find a unit tangent vector T to the curve.
(b) If r is the position vector of a particle moving on C at time t, verify in this
case that v = vT.
(a) A tangent vector to C is
dr/dt =  6 sin 2* i + 6 cos 2t j + 8k
The magnitude of this vector is
\dr/dt\ = ds/dt = V( 6 sin 2t)* + (6 cos 2*)* + (8)* = 10
Then a unit tangent vector to C is
dr/dt dx/dt dt  6 sin 2* i 4 6 cos 2t j + 8k
T =
dx/dt ds/dt ds 10
=  1 sin 2t i + f cos 2t j + $ k
(6) This follows at once from (a) since
v = dr/dt =  6 sin 2t i + 6 cos 2t j + 8k
= (10)( f sin 2* i + f cos2t j + $k) = vT
Note that in this case the speed of the particle along the curve is constant.
1.33. If T is a unit tangent vector to a space curve C, show that dT/ds is normal to T.
Since T is a unit vector, we have T*T = 1. Then differentiating with respect to 8, we obtain
T .f + *. T = 2T M = o or T.£=
d8 da d8 da
which states that dT/ds is normal, i.e. perpendicular, to T.
20 VECTORS, VELOCITY AND ACCELERATION [CHAP. 1
If N is a unit vector in the direction of dT/ds, we have
dT/ds = kS
and we call N the unit principal normal to C. The scalar k = \dT/ds\ is called the curvature,
while R = 1/k is called the radius of curvature.
1.34. Find the (a) curvature, (b) radius of curvature and (c) unit principal normal N to
any point of the space curve of Problem 1.32.
(a) From Problem 1.32, T = f sin It i + § cos 2t j + k. Then
dT _ dT/dt _ (6/5) cos 2t i  (6/5) sin 2t j
ds ds/dt 10
= — Jj cos 2t i — ^ sin It j
Thus the curvature is c =
= V(^cos 2 *) 2 + (^sin2«)2 = £
ds
(b) Radius of curvature = R = 1/ K = 25/3
(c) From (a), (6) and Problem 1.33,
., 1 dT n dT _ x . . „ .
N = KdF = ^rfT = cos2*ism2«j
1.35. Show that the acceleration a of a particle which travels along a space curve with
velocity v is given by , 9
a = 3t T +ff N
where T is the unit tangent vector to the space curve, N is its unit principal normal
and R is the radius of curvature.
Velocity v = magnitude of v multiplied by unit tangent vector T, or
v = vT
Differentiating, a = % = * {vT) = ^ T + ^
dt dt dt dt
But ^ = £?I ds_ ds = tjN
dt ds dt KXN dt KVS * R
ti.«« — ^ V m . /vfi\ dv _ , i;2 T
Then a  * T + *(«; = d* T + b n
This shows that the component of the acceleration is dv/dt in a direction tangent to the path and
v 2 /R in the direction of the principal normal to the path. The latter acceleration is often called
the centripetal acceleration or briefly normal acceleration.
CIRCULAR MOTION
1.36. A particle moves so that its position vector is given by r = cos o>t i + sin U j where
o> is a constant. Show that (a) the velocity v of the particle is perpendicular to r,
(b) the acceleration a is directed toward the origin and has magnitude proportional
to the distance from the origin, (c) r x v = a constant vector.
/ \ dr
di ~ ~ a sm ut * "*" " cos w * *' Then
r • v = [cos at i + sin at j] • [— « sin at i + w cos at j]
= (cos at)(—a sin at) + (sin at)(a cos at) =
and r and v are perpendicular.
CHAP. 1]
VECTORS, VELOCITY AND ACCELERATION
21
(b) ^ = ^ = a 2 cos at i  a 2 sin «t j = w 2 [cos <ot i + sin at j] = « 2 r
7 at 2 at
Then the acceleration is opposite to the direction of r, i.e. it is directed toward the origin.
Its magnitude is proportional to r which is the distance from the origin.
(c) r X v = [cos at i + sin at j] X [— a sin at i + a cos at j]
i j k
cos at sin at
— a sin at a COS at
Physically, the motion is that of a particle moving on the circumference of a circle with
constant angular speed a. The acceleration, directed toward the center of the circle, is the
centripetal acceleration.
= «(cos 2 at + sin 2 at)k = «k, a constant vector.
GRADIENT, DIVERGENCE AND CURL
1.37. If = x 2 yz* and A = xziy 2 j + 2x 2 yk, find (a) V</>, (&) V'A, (c) V x A, (d) div
(<j>A), (e) curl (<f>A).
1 ' v \3« dy ' dz }* dx dy* dz
= ^(x2yz3)i + £(x*yz3)j + ^(as»y*»)k = 2<ci/z 3 i + ****j + 3x*y**k
(6) VA = (£l + £j + £k)<^lrt + *«»*>
(c)
VXA = (^ i + ^ + ^ k ) X( ^ i " 2/2J +
2x 2 yk)
i j k
3/fla; d/dy d/dz
xz —y 2 2x^y
= 2x 2 i + (x — 4*2/) j
(d) div (*A) = V • (*A) = V • (x^yzH  x*yHH + 2* 4 i/ 2 z 3 k)
+ f(2a*2/ 2 z 3 )
oz
(e) curl (*A) = V X (*A) = V X (x*yz*i  *WJ + 2* 4 i/ 2 z 3 k)
i j k
d/dx d/dy d/dz
xSyz* x 2 y*z* 2ar% 2 « 8
= (4x*yz s + 3« 2 i/ 3 « 2 )i + (4«3j/2 3  8x 3 i/ 2 z 3 )j  (2xy s z* + o 3 « 4 )k
1.38. (a) If A = (2xy + z?)i + (x 2 + 2y)j+{Sxz 2 2)k, show that VXA
(b) Find a scalar function <j> such that A = V«/>
i j k
(a) VXA = d/dx d/dy d/dz =
2xy + « 3 a 2 + 2y 3s« 2  2
= 0.
22 VECTORS, VELOCITY AND ACCELERATION [CHAP. 1
(b) Method 1. If A = V<& = ^i + f2j + ^k then we must have
ox dy dz
(1) f£ = 2xy + * 3 (*) £ = x2 + 2y (3) '& = 3xz2  2
da; fly flz
Integrating, we find
U) 4> = sfiy + xtfl + F^x) (5) * = x 2 y + y 2 + F 2 (x,z)
(6) <f» = xz*2z + F 3 (x,y)
Comparing these we must have F^y^) = y 2 — 2z, F 2 (x,z) = xz* — 2z, F z (x,y) = x*y + y*
and so <f> = x 2 y + xz s + y 2 — 2z.
Method 2. We have if A = v>,
Adr = ^i + ^j + ^k^.^i + ^j + ^k)
= H dx + % dy + fz dz = d *
an exact differential. For this case,
d<f> = A'dr = (2xy + z*)dx + (x 2 + 2y)dy + (Sxz 2 2)dz
= [(2xy + z s ) dx + x 2 dy + Sxz 2 dz] + 2y dy  2 dz
= d(x 2 y + xz*) + d(y 2 ) + d(2z)
 d(x 2 y + xz* + y 2  2z)
Then <p = x 2 y + xz s + y 2 — 2z. Note that an arbitrary constant can also be added to <f>.
LINE INTEGRALS AND INDEPENDENCE OF THE PATH
1.39. If A = (Sx 2  6yz)i + (2y + 3xz)j + (1  4xyz 2 )k, evaluate J A'dr from (0,0,0) to
(1, 1, 1) along the following paths C: c
(a) x = t, y = t 2 , zi^.
(b) the straight lines from (0, 0, 0) to (0, 0, 1), then to (0, 1, 1), and then to (1, 1, 1).
(c) the straight line joining (0,0,0) and (1,1,1).
f A • dr = f {(Sx 2  6yz)i + (2y + 3xz)j + (1  4xyz 2 )k} '(dxi + dyj + dz k)
= I (Sx 2 6yz)dx + (2y + Sxz)dy + (l4xyz 2 )dz
(a) If x = t, y = t 2 , z = «*, points (0, 0, 0) and (1, 1, 1) correspond to t = and t = 1 respectively. Then
f Adr = f {3*2  6(*2)(*3)} dt + {2*2 + 3(*)(*3)} d(t 2 ) + {1  4(*)(*2)(*3)2} d(*»)
J o J t=0
f (3*26* 5 )d* + (4t? + 6t*)dt + (St 2  12* 11 ) dt = 2
J t=o
Another method.
Along C, A = (3*2  6*5)i + (2*2 + 3*4)j + (i _ 4«9) k an d r = »i + yj + «k = *i + * 2 j + *3fc,
dx = (i + 2<j + 3< 2 k) dt. Then
J A'dr = J (3*2
6*5) dt + (4*8 + 6*5) dt + (St 2  12*") dt = 2
(6) Along the straight line from (0, 0, 0) to (0, 0, 1), x — 0, y = 0, dx = 0, dy = while z varies from
to 1. Then the integral over this part of the path is
f {3(0) 2 6(0)(z)}0 + {2(0) + 3(0)(«)}0 + {1  4(0)(0)(z 2 )} dz = f dz = 1
CHAP. 1]
VECTORS, VELOCITY AND ACCELERATION
23
Along the straight line from (0, 0, 1) to (0, 1, 1), x = 0, z = 1, dx = 0,dz = while y varies
from to 1. Then the integral over this part of the path is
2ydy = 1
l r 1
f {3(0) 2  6(y){l)}0 + {2i/ + 3(0)(1)} dy + {1  4(0)(i/)(l) 2 }0 = I
J y =o ^y=o
Along the straight line from (0, 1, 1) to (1, 1, 1), y = l,z = 1, dy = 0,dz = while x varies
from to 1. Then the integral over this part of the path is
C {3*2  6(1)(1)} dx + {2(1) + 3z(l)}0 + {1  4*(1)(1) 2 >0 = f (3* 2  6) dx = 5
x=0
Adding, I
Jr.
A • dr = 1 + 15 = 3.
(c) Along the straight line joining (0,0,0) and (1,1,1) we have x = t, y = t, z = t. Then since
dx = dy = dz = dt,
( A'dr = f (3x 2 6yz)dx + (2y + Sxz) dy + (1  4xyz 2 ) dz
= f (3t 2 6t*)dt + (2t + St 2 )dt + (l4t*)d«
f (2« + l4t*)dt = 6/5
Note that in this case the value of the integral depends on the particular path.
1.40. If A = (2xy + z?)i + (x 2 + 2y)j + (3zz 2 2)k show that (a) J A dr is independent
of the path C joining the points (1,1,1) and (2,1,2) and (b) find its value.
By Problem 1.38, V X A = or A • dr = d<p = d(x 2 y + xz* + y 2  2z). Then the integral is
independent of the path aijd its value is
J (2,1,2) /•(2.1.2)
A'dr = I d(x*y + xz 3 + y 2  2c)
ii.i.i) *^(i,i.D
= x 2 y + xz* + /222
(2.1,2)
(1,1.1)
= 18
MISCELLANEOUS PROBLEMS
1.41. Prove that if a and b are noncollinear, then xa + yb = implies x  y = 0.
Suppose x ¥= 0. Then a?a + yb = implies xa = — yb or a = —(y/x)b, i.e. a and b must be parallel
to the same line (collinear), contrary to hypothesis. Thus x = 0; then 2/b = 0, from which y = 0.
1.42. Prove that the diagonals of a parallelogram bi
sect each other.
Let ABCD be the given parallelogram with diagonals
intersecting at P as shown in Fig. 128.
Since BD + a = b, BD = b  a. Then BP = x(b  a).
Since AC = a + b, AP = y(a + b).
But AB = AP + PB = AP  BP,
i.e. a = i/(a + b) — x(b — a) = (x + y)a + (y — x)b.
Since a and b are noncollinear we have by Problem
1.41, x + y = 1 and y — x = 0, i.e. x = y = % and P
is the midpoint of both diagonals.
24
VECTORS, VELOCITY AND ACCELERATION
[CHAP. 1
1.43. Prove that for any vector A,
(a) A = (Ai)i + (Aj)j + (Ak)k
(b) A = A(cos a i + cos /? j + cos y k)
where a, /?, y are the angles which A makes with i, j, k respectively and cos a, cos (3,
cos y are called the direction cosines of A.
(a) We have A = A t i + A 2 j + A 3 k. Then
A«i = (A x i + A 2 j + A 3 k) • i = A x
A»j = (Aji + Aaj + Agk)'! = A 2
A'k = (Aii + AaJ + AgkJk = A 3
Thus
A = (A • i)i + (A • j)j + (A • k)k
(b)
A • i = A i cos a = A cos a
A«j — Ajcos/? = Acos/3
A»k = A k cos y = A cos y
Then from part (a),
A = (A • i)i + (A • j)j + (A • k)k = A(cosa i + cos/3 j + cosy k)
1.44. Prove that V<£ is a vector perpendicular to the surface </>(#, 2/, 2) = c, where c is a
constant.
Let r = xi + j/j + zk be the position vector to any point P(x, y, z) on the surface.
Then dv — dx i + dy j + dz k lies in the plane tangent to the surface at P. But
* = ^ dx + ^ dy+ ^ dz
or
i.e. V# • dr = so that V^ is perpendicular to dr and therefore to the surface
ff i + lf i+ lf k » (&i + < "' i + &k)
1.45. Find a unit normal to the surface 2# 2 + 4yz  5z 2 = 10 at the point P(3,l,2).
By Problem 1.44, a vector normal to the surface is
V(2ar2 + Ayz  5z 2 ) = 4ai + 4zj + (Ay  10z)k = 12i + 8j  24k at (3,1,2)
12i + 8j  24k 3i + 2j  6k
Then a unit normal to the surface at P is
Another unit normal to the surface at P is
V(12) 2 + (8) 2 + (24)2
3i + 2j  6k
1.46. A ladder AB of length a rests against a vertical wall OA [Fig. 129]. The foot B
of the ladder is pulled away with constant speed vo. {a) Show that the midpoint
of the ladder describes the arc of a circle of radius a/2 with center at O. (b) Find
the velocity and speed of the midpoint of the ladder at the instant where B is distant
b < a from the wall.
(a) Let r be the position vector of midpoint M of AB.
If angle OB A = o, we have
OB = a cos e i, OA = a sin e j
AB = OB — OA = a cos i — a sin j
Then
r = OA + AM = OA + £AB
= a sin e j + \(a> cos e i — a sin e j)
= ^a(cos 9 i + sin e j)
Thus r = ^a, which is a circle of radius a/2
with center at 0. Fig. 129
CHAP. 1]
VECTORS, VELOCITY AND ACCELERATION
25
(6) The velocity of the midpoint M is
jr = jrtya(co& 8 i + sin e j)} = ^a(— sin e ei + cos 8 8j)
where 8 = d8/dt.
The velocity of the foot B of the ladder is
vA = r(OB) = 37 (a cos 0i) = — a&m8 8i or a sin 8 8 =
at at
At the instant where B is distant 6 from the wall we have from (2),
v
am 8 =
Va 2  62
8 =
v
n
a a sin* y a 2 _ b 2
Thus from (1) the required velocity of M at this instant is
dr 1 / .
y/cPb*
and its speed is av„/2 Va 2 — o 2 .
(i)
(«)
1.47. Let (r, 0) represent the polar coordinates describing the position of a particle. If
ri is a unit vector in the direction of the position vector r and $i is a unit vector
perpendicular to r and in the direction of increasing 6 [see Fig. 130], show that
(a) ri = cos 9 i 4 sin 6 j, 0i = — sin 6 i + cos j
(6) i = cos ri — sin 0i, j = sin ri + cos 6 $i
(a) If r is the position vector of the particle at any
time t, then dr/dr is a vector tangent to the
curve 8 = constant, i.e. a vector in the direc
tion of r (increasing r). A unit vector in this
direction is thus given by
_ 5r /Idrl
dr/ \dr\
(1)
Since
r = xi + yj = rcosfli + rsintfj
as seen from Fig. 130, we have
3r . , . . \dr\
— = cos* i + sin j, — = 1
dr \dr\
so that
cos 8 i + sin 8 j
(3)
Fig. 130
Similarly, Br/d8 is a vector tangent to the curve r = constant. A unit vector in this
direction is thus given by
Now from (2),
dr ... . 3r
— = — r sin 8i + rcosff j, — = r
do * \de
so that (4) yields
8 X = — sin 8 i + cos 8 j (5)
(b) These results follow by solving the simultaneous equations (3) and (5) for i and j.
26 VECTORS, VELOCITY AND ACCELERATION [CHAP. 1
1.48. Prove that (a)r a = 00i (b)di = 0r u
(a) Prom (S) of Problem 1.47 we have
Tl dt dr dt de dt
= (0)(f ) + ( sin e i + cos e })($) = e0 t
(b) From (5) of Problem 1.47 we have
_ d*i _ d*idr , ^ide_
01 ~ dt ~ dr dt de dt
= (0)(r) + (— cos e i — sin e j)(e) = — er x
1.49. Prove that in polar coordinates (a) the velocity is given by
v = rri 4 r$6i
and (b) the acceleration is given by
a = ( r  r6 2 )ri + (r 6 + 2r9) 9 1
(a) We have r = rr 1 so that
dr dr , ^ r i ... . .
v = Tt = Tt r * + r dT = ^ + rr ^ = rr i + *"i
by Problem 1.48(a).
(6) From part (a) and Problem 1.48 we have
dv d .* , • x
a = Tt = de (rr ^ + r ^ l)
= rr t + rr x + r0«x + rtf*! + rebi
= Vr! + f(5#i) + ffoi + r'e'9 1 + (re^—eri)
= (r — r$ 2 )r t + (r6>'+2r£)0 1
Supplementary Problems
VECTOR ALGEBRA
1.50. Given any two vectors A and B, illustrate geometrically the equality 4A+3(BA) = A + 3B.
1.51. Given vectors A, B and C, construct the vectors (a) 2A  3B + £C, (6) C  £A + JB.
1.52. If A and B are any two nonzero vectors which do not have the same direction, prove that pA + qB
is a vector lying in the plane determined by A and B.
1.53. (a) Determine the vector having initial point (2,1,3) and terminal point (3,2,4). (6) Find the
distance between the two points in (a). Ans. (a) i + 3j — 7k, (6) \/59
1.54. A triangle has vertices at the points A(2,l,1), £(1,3,2), C(l,2,1). Find the length of the
median to the side AB. Ans. £\/66
1.55. A man travels 25 miles northeast, 15 miles due east and 10 miles due south. By using an
appropriate scale determine (a) graphically and (6) analytically how far and in what direction
he is from his starting position. Ans. 33.6 miles, 13.2° north of east.
CHAP. 1] VECTORS, VELOCITY AND ACCELERATION 27
1.56. Find a unit vector in the direction of the resultant of vectors A = 2i — j + k, B = i + j + 2k,
C = 3i2j + 4k. Ans. ±(6121 + 7^/^
THE DOT OR SCALAR PRODUCT
1.57. Evaluate (A + B) • (AB) if A = 2i3j + 5k and B = 3i + j2k. Ans. 24
1.58. Find a so that 2i — 3j + 5k and 3i + aj — 2k are perpendicular. Ans. a = —4/3
1.59. If A = 2i + j + k, B = i  2j + 2k and C = 3i  4j + 2k, find the projection of A + C in the
direction of B. Ans. 17/3
1.60. A triangle has vertices at A(2,3,l), B(l, 1,2), C(l,2,3). Find the acute angle which the
median to side AC makes with side BC. Ans. cos 1 V91/14
1.61. Prove the law of cosines for triangle ABC, i.e. c 2 = a 2 + b 2 — 2ab cos C.
[Hint. Take the sides as A,B,C where C = A — B. Then use OC = (A — B) • (A — B).]
1.62. Prove that the diagonals of a rhombus are perpendicular to each other.
THE CROSS OR VECTOR PRODUCT
1.63. If A = 2ij + k and B = i + 2j3k, find (2A + B) X (A2B). Ans. 2byfl
1.64. Find a unit vector perpendicular to the plane of the vectors A = 3i — 2j + 4k and B = i + j — 2k.
Ans. ±(2j + k)/V5
1.65. Find the area of the triangle with vertices (2,3,1), (1,1,2), (1,2,3). Ans. \yf%
1.66. Find the shortest distance from the point (3,2,1) to the plane determined by (1,1,0), (3,1,1),
(1,0,2). Ans. 2
* /.ft t> 0.1.7 * • j. x • i a ns, • sin A sin B sin C
1.67. Prove the law of sines for triangle ABC, i.e. = — r — = •
a o c
[Hint. Consider the sides to be A, B, C where A + B + C = and take the cross product of both
sides with A and B respectively.]
TRIPLE PRODUCTS
1.68. If A = 2i + j  3k, B = i  2j + k and C = i + j  4k, find (a) A • (B X C), (b) C • (A X B),
(c) A X (B X C), (d) (A X B) X C. Ans. (a) 20, (6) 20, (c) 8i  19j  k, {d) 25i  15j  10k
1.69. Prove that A • (B X C) = (A X B) • C, i.e. the dot and the cross can be interchanged.
1.70. Find the volume of a parallelepiped whose edges are given by A = 2i + 3j — k, B = i — 2j + 2k,
C = 3ij2k. Ans. 31
1.71. Find the volume of the tetrahedron with vertices at (2, 1, 1), (1, — 1, 2), (0, 1, 1), (1„ 2, 1).
Ans. 4/3
1.72. Prove that (a) A(BXC) = B«(CXA) =a C • (A X B),
(b) A X (B X C) = B(A • C)  C(A • B).
1.73. (a) Let r v r 2 , r 3 be position vectors to three points P 1 ,P 2 ,P3 respectively. Prove that the equation
(r — rj) • [(r — r 2 ) X (r — r 3 )] = 0, where r = xi + yj + zk, represents an equation for the plane
determined by P u P 2 and P 3 . (&) Find an equation for the plane passing through (2,1,2),
(1,2,3), (4,1,0). Ans. (b) 2x + y3z = 9
DERIVATIVES AND INTEGRALS OF VECTORS
1.74. Let A = 3fi  (t 2 + t)j + (t*  2t 2 )k. Find (a) dA/dt and (6) cPA/dt 2 at t = 1.
Ans. (a) 3i3jk, (6) 2j + 2k
1.75. If r = a cos ut + b sin at, where a and b are any constant noncollinear vectors and « is a constant
scalar, prove that (a) r X dr/dt = <o(a X b), (6) d 2 r/dt 2 + <o 2 r = 0.
28 VECTORS, VELOCITY AND ACCELERATION [CHAP. 1
1.76. If A = t\ — sin t k and B = cos t i + sin t j + k, find ^(A«B). Ans. — t sin *
1.77. Prove that t(AXB) = AXr + rXB where A and B are differentiate functions of u.
du ' du du
1.78. If A(m) = 4(tt  l)i  (2u + 3)j + Gu 2 k, evaluate (a) J A(u) du, (b) J (ui  2k) • A(u) du.
Ans. (a) 6i  8j + 38k, (b) 28 2 *
1.79. Find the vector B(u) such that d 2 B/du 2 = Gul  48w 2 j + 12k where B = 2i  3k and dB/du =
i + 5j for u = 0. Ans. (u s + u + 2)i + (5u  4u 4 )j + (Gu 2  3)k
1.80.
Xd?A dA
A X j^ dt = A X rr + c where c is a constant vector.
1.81. If R = x 2 yi  2y 2 zj + xy 2 z>k, find
3 2 B fl 2 R
aa; 2 3j/ 2
at the point (2,1,2). Ans. IGyfh
3 2
1.82. If A = x*i  y] + xzk and B = yi + xjxyzk, find ^y(AXB) at the point (1,1,2).
Ans. — 4i + 8j
VELOCITY AND ACCELERATION
1.83. A particle moves along the space curve r = (t 2 + *)i + (8*  2)j + (2t»  4t 2 )k. Find the
(a) velocity, (6) acceleration, (c) speed or magnitude of velocity and (d) magnitude of accelera
tion at time t = 2. Ans. (a) 5i + 3j + 8k, (6) 21 + 16k, (c) ly[2, (d) 2^
1.84. A particle moves along the space curve defined by x = e* cost, y — e~ t sin t, z = e~ t . Find
the magnitude of the (a) velocity and (6) acceleration at any time t.
Ans. (a) V3~e _t , (6) VEe*
1 .85. The position vector of a particle is given at any time t by r = a cos at i + 6 sin ut j + ct 2 k.
(a) Show that although the speed of the particle increases with time the magnitude of the
acceleration is always constant. (&) Describe the motion of the particle geometrically.
RELATIVE VELOCITY AND ACCELERATION
1.86. The position vectors of two particles are given respectively by r x = ti — 1 2 \ + (2t + 3)k and
r 2 = (2t  3« 2 )i + 4tj  £%. Find (a) the relative velocity and (&) the relative acceleration of the
second particle with respect to the first at t = l. Ans. (a) 5i + 6j5k, (6) 6i + 2j6k
1.87. An automobile driver traveling northeast at 26 mi/hr notices that the wind appears to be coming
from the northwest. When he drives southeast at 30 mi/hr the wind appears to be coming from
60° south of west. Find the velocity of the wind relative to the ground.
Ans. 52 mi/hr in a direction from 30° south of west
1.88. A man in a boat on one side of a river wishes to reach a point directly opposite him on the other
side of the river. Assuming that the width of the river is D and that the speeds of the boat and
current are V and v < V respectively, show that (a) he should start his boat ups tream at an angle
of sin 1 (v/V) with the shore and (6) the time to cross the river is D/y/V 2  v 2 .
TANGENTIAL AND NORMAL ACCELERATION
1.89. Show that the tangential and normal acceleration of a particle moving on a space curve are given
by d 2 s/dt 2 and K (ds/dt) 2 where s is the arc length of the curve measured from some initial point
and k is the curvature.
1.90. Find the (a) unit tangent T, (6) principal normal N, (c) radius of curvature R and (d) curvature
k to the space curve x = t, y = t 2 /2, z = t.
Ans. (a) (i+tj + k)/v^T2, (&) (ti + 2j  tk)/VW+4, (c) (t 2 + 2f' 2 lyf2, (d) V2/(* 2 + 2) 3 ' 2
CHAP. 1] VECTORS, VELOCITY AND ACCELERATION 29
1.91. A particle moves in such a way that its position vector at any time t is r = £i + ^ 2 j + ik.
Find (a) the velocity, (6) the speed, (c) the acceleration, (d) the magnitude of the acceleration,
(e) the magnitude of the tangential acceleration, (/) the magnitude of the normal acceleration.
Ans. (a) i + *j + k, (6) y/W+2, (c) j, (d) 1, (e) t/V& + 2, (/) V2/V^T2
1.92. Find the (a) tangential acceleration and (6) normal acceleration of a particle which moves on
the ellipse r = a cos at i + b sin at j.
M 2( a 2 _ fr2) sm at cos at a 2 ab
An8. (a) , (6) .
Va 2 sin 2 at + b 2 cos 2 u£ V a 2 sin 2 <ot + 6 2 cos 2 at
CIRCULAR MOTION
1.93. A particle moves in a circle of radius 20 cm. If its tangential speed is 40 cm/sec, find (a) its
angular speed, (6) its angular acceleration, (c) its normal acceleration.
Ans. (a) 2 radians/sec, (6) radians/sec 2 , (c) 80 cm/sec 2
1.94. A particle moving on a circle of radius R has a constant angular acceleration a. If the particle
starts from rest, show that after time t (a) its angular velocity is a — at, (b) the arc length
covered is 8 = ^Rat 2 .
1.95. A particle moves on a circle of radius R with constant angular speed « . At time t = it starts
to slow down so that its angular acceleration is —a (or deceleration a). Show that (a) it comes to
rest after a time u /a and (6) has travelled a distance Raf,/2a.
1.96. If the particle in Problem 1.95 is travelling at 3600 revolutions per minute in a circle of radius
100 cm and develops a constant deceleration of 5 radians/sec 2 , (a) how long will it be before it
comes to rest and (6) what distance will it have travelled? Ans. (a) 75.4 sec, (6) 1.42 X 10 e cm
GRADIENT, DIVERGENCE AND CURL
1.97. If A = xzi+(2x 2 y)jyz 2 k and <f> = $x 2 y + yW, find (a) V^, (6) V»A and (c) VXA
at the point (1,1,1). Ans. (a) 6i + j + 3k, (6) 2, (c) i + j + 4k
1.98. If <p = xy + yz + zx and A = x 2 y\ + yH\ + z 2 xk, find (a)A«V^, (6)^V»A and (c) (V#) X A
at the point (3, 1, 2). Ans. (a) 25, (b) 2, (c) 56i  30j + 47k
1.99. Prove that if U,V,A,B have continuous partial derivatives, then (o) V(17 + V) = V17+VV,
(b) V • (A + B) = V • A + V • B, (c) V X (A + B) = V X A + V X B.
1.100. Show that V X (r 2 r) = where r = xi + yj + zk and r = r.
1.101. Prove that (a) div curl A = and (6) curl grad <j> — under suitable conditions on A and <p.
1.102. If A = (2x 2 yz)i + (y 2 2xz)j + x 2 z*k and <p = x 2 y  Sxz 2 + 2xyz, show directly that
div curl A = and curl grad $ — 0.
1.103. If A = 3»z 2 i  yzj + (x + 2z)k, find curl curl A. Ans. 6xi + (6*l)k
1.104. (a) Prove that V X (V X A) = V 2 A+ V(V • A), (b) Verify the result in (a) if A is as given in
Problem 1.103.
1.105. Prove: (a) V X (UA) = (VC7) X A+ U{V X A). (6) V • (AX B) = B« (V X A)  A' (V X B).
LINE INTEGRALS AND INDEPENDENCE OF THE PATH
1.106. If F = (3aj  2y)i + (y + 2z)j  x 2 k, evaluate \ F > dr from (0, 0, 0) to (1, 1, 1), where C is a path
consisting of: (a) the curve x = t, y = t 2 , z = £ 3 ; (6) a straight line joining these points; (c) the
straight lines from (0,0,0) to (0,1,0), then to (0,1,1) and then to (1,1,1); (d) the curve x = z 2 ,
z = y 2 . Ans. (a) 23/15, (6) 5/3, (c) 0, (d) 13/30
30 VECTORS, VELOCITY AND ACCELERATION [CHAP. 1
1.107. Evaluate j Adr where A = 3x 2 i K (2xz — y) j + zk along (a) the straight line from (0,0,0)
to (2,1,3), (b) the space curve x = 2t 2 , y — t, z — 4t 2 — t from t = to t = l, (c) the curve defined
by x 2 = 4y, 3x* = 8z from x = to x = 2. Ans. (a) 16, (6) 14.2, (c) 16
1.108. Find 4> F»dr where F = (a; — 3#)i + (y — 2x)i and C is the closed curve in the xy plane,
x = 2 cos t, i/ = 3 sin t, z = from t = to t = 2ir. Ans. 6tt
1.109. (a) If A = {Axy  Zx 2 z 2 )i + (4y + 2ar 2 )j + (1  2x s z)k, prove that I A'dr is independent of the
curve C joining two given points. (6) Evaluate the integral in (a) if C is the curve from the
points (1,1,1) to (2,2,1). Ans. (b) 19
1.110. Determine whether I A • dr is independent of the path C joining any two points if (a) A = 2xyzi +
J c
x 2 zj + x 2 yk, (b) 2xzi + (x 2 — y)j + (2z — x 2 )k. In the case where it is independent of the path,
determine <j> such that A = V#.
Ans. (a) Independent of path, <f> = x 2 yz + c; (b) dependent on path
1.111. Evaluate (J) E • dr where E = rr. Ans.
MISCELLANEOUS PROBLEMS
1.112. If A X B = 8i  14j + k and A + B = 5i + 3j + 2k, find A and B.
Ans. A = 2i + j2k, B = 3i + 2j + 4k
1.113. Let l^m^n! and Z 2 >™2« w 2 be direction cosines of two vectors. Show that the angle between them
is such that cos e = IJ2 + m 1 m 2 + n 1 % 2 «
1.114. Prove that the line joining the midpoints of two sides of a triangle is parallel to the third side
and has half its length.
1.115. Prove that (A X B) 2 + (A • B) 2 = A 2 B 2 .
1.116. If A, B and C are noncoplanar vectors [vectors which do not all lie in the same plane] and
x t A + y{B + «]C = x 2 A + y 2 B + z 2 C, prove that necessarily x x — x 2 , y\ — y%, z x — z 2 .
1.117. Let ABCD be any quadrilateral and points P, Q, R and S the midpoints of successive sides. Prove
that (a) PQRS is a parallelogram, (6) the perimeter of PQRS is equal to the sum of the lengths
of the diagonals of ABCD.
1.118. Prove that an angle inscribed in a semicircle is a right angle.
1.119. Find a unit normal to the surface x 2 y  2xz + 2y 2 z^ = 10 at the point (2, 1, 1).
Ans. ±(Zi + 4}6k)/^
 <„» ^ m ^ . dA . dA
1.120. Prove that A • ^ = A ^ .
1.121. If A(w) is a differentiate function of u and A(te) = 1, prove that dA/du is perpendicular to A.
1.122. Prove V(*A) = (V*) • A + <f>(V • A).
1.123. If A X B = A X C, does B = C necessarily? Explain.
1.124. A ship is traveling northeast at 15 miles per hour. A man on this ship observes that another ship
located 5 miles west seems to be traveling south at 5 miles per hour, (a) What is the actual
velocity of this ship? (6) At what distance will the two ships be closest together?
1.125. Prove that (A X B) • (C X D) + (B X C) • (A X D) + (C X A) • (B X D) = 0.
1.126. Solve the equation d 2 r/dt 2 = gk where g is a constant, given that r = 0, dr/dt = vjs. at t = 0.
Ans. r = (v t — ^gt 2 )k
CHAP. 1] VECTORS, VELOCITY AND ACCELERATION 31
1.127. If <t> = (« 2 + 2/ 2 + 2 2 ) 1/2 , show that V 2 ^ = V • (Vtf>) = at all points except (0,0,0).
1.128. The muzzle velocity of a gun is 60 mi/hr. How long does it take for a bullet to travel through the
gun barrel which is 2.2 ft long, assuming that the bullet is uniformly accelerated? Ans. .05 sec
1.129. A 25 foot ladder AB rests against a vertical wall OA as in Fig. 129, page 24. If the foot of the
ladder B is pulled away from the wall at 12 ft/sec, find (a) the velocity and (6) the acceleration
of the top of the ladder A at the instant where B is 15 ft from the wall.
Ans. (a) 9 ft/sec downward, (6) 11.25 ft/sec 2 downward
1.130. Prove that (a)A + B^ A + B, (6) A+B + C ^ A + B + C. Give a possible geometric
interpretation.
1.131. A train starts from rest with uniform acceleration. After 10 seconds it has a speed of 20 mi/hr.
(a) How far has it traveled from its starting point after 15 seconds and (6) what will be its speed
in mi/hr? Ans. (a) 330 ft, (6) 30 mi/hr
1.132. Prove that the magnitude of the acceleration of a particle moving on a space curve is
y/(dvldt)* + vVR 2
where v is the tangential speed and R is the radius of curvature.
1.133. If T is the unit tangent vector to a curve C and A is a vector field, prove that
JA'dr = \ A'Tds
c J c
where s is the arc length parameter.
1.134. If A = (2x  y + 4)i + (5y + Sx  6)j, evaluate 4> A'dr around a triangle with vertices at
(0,0,0), (3,0,0), (3,2,0). Ans. 12 J
1.135. An automobile driver starts at point A of a highway and stops at point B after traveling the
distance D in time T. During the course of the trip he travels at a maximum speed V. Assuming
that the acceleration is constant both at the beginning and end of the trip, show that the time
during which he travels at the maximum speed is given by 2D/V — T.
1.136. Prove that the medians of a triangle (a) can form a triangle, (b) meet in a point which divides the
length of each median in the ratio two to one.
1.137. If a particle has velocity v and acceleration a along a space curve, prove that the radius of
curvature of its path is given numerically by
R = ^r
vXa
1.138. Prove that the area of a triangle formed by vectors A, B and Cis £AXB + BXC + CXA.
1.139. (a) Prove that the equation AXX = B can be solved for X if and only if A»B = and A^0.
(6) Show that one solution is X = B X A/A 2 , (c) Can you find the general solution?
Ans. (c) X = B x A/A 2 + XA where X is any scalar.
1.140. Find all vectors X such that A • X = p.
Ans. X = pA/A 2 + V X A where V is an arbitrary vector.
1.141. Through any point inside a triangle three lines are constructed parallel respectively to each of
the three sides of the triangle and terminating in the other two sides. Prove that the sum of the
ratios of the lengths of these lines to the corresponding sides is 2.
1.142. If T, N and B = T X N are the unit tangent vector, unit principal normal and unit binormal to
a space curve r = r(u), assumed differentiate, prove that
dT dB *t < * n «i'
Is = kN ' * = ~ tN ' Is  tB ~ kT
These are called the FrenetSerret formulas. In these formulas k is called the curvature, r is the
torsion and their reciprocals R = 1/k, o = Vt are called the radius of curvature and radius of
torsion.
32
VECTORS, VELOCITY AND ACCELERATION
[CHAP. 1
1.143. In Fig. 131, AB is a piston rod of length I. If A moves along horizontal line CD while B moves
with constant angular speed w around the circle of radius a with center at O, find (a) the velocity
and (6) the acceleration of A.
P
P
lS
l /
Fig. 131
Fig. 132
1.144. A boat leaves point P [see Fig. 132] on one side of a river bank and travels with constant velocity
V in a direction toward point Q on the other side of the river directly opposite P and distance D
from it. If r is the instantaneous distance from Q to the boat, is the angle between r and PQ,
and the river travels with speed v, prove that the path of the boat is given by
_ D sec 6
~ (sec e + tan e) v/v
1.145. If v = V in Problem 1.144, prove that the path is an arc of a parabola.
L146. (a) Prove that in cylindrical coordinates (p,<j>,z) [see Fig. 133] the position vector is
r = p cos <t> i + p sin <f> j + zk
(6) Express the velocity in cylindrical coordinates,
(c) Express the acceleration in cylindrical coordinates.
Ans. (6) v = ppx 4 p$$\ + «k
(c) a = (p P* 2 )pi + (p'4+ 2p£)tfi + *k
Illlplllilplllliplim
2
*
p'v.
Cylindrical coordinates
Fig. 133
Spherical coordinates
Fig. 134
1.147. (a) Prove that in spherical coordinates (r, $, <j>) [see Fig. 134] the position vector is
r = r sin e cos <f> i + r sin 9 sin # j + r cos k
(6) Express the velocity in spherical coordinates,
(c) Express the acceleration in spherical coordinates.
Ana. (6) v = rt x + rb§ x + rj> sin $ f t
(c) a = (V  rff 2 — r£ 2 sin 2 tf)r x + (2re + rV — r^ 2 sin cos *)«!
+ (2re$ + 2r# sin 6 + r'<p sin 0)# x
1.148. Show that if a particle moves in the xy plane the results of Problems 1.146 and 1.147 reduce to
those of Problem 1.49.
WORK, ENERGY and MOMENTUM
NEWTON'S LAWS
The following three laws of motion given by Sir Isaac Newton are considered the
axioms of mechanics:
1. Every particle persists in a state of rest or of uniform motion in a straight line
(i.e. with constant velocity) unless acted upon by a force.
2. If F is the (external) force acting on a particle of mass m which as a consequence is
moving with velocity v, then
F = 5<""> = % «
where p = mv is called the momentum. If m is independent of time t this becomes
dx
F = mjT = ma (#)
where a is the acceleration of the particle.
3. If particle 1 acts on particle 2 with a force Fi 2 in a direction along the line joining
the particles, while particle 2 acts on particle 1 with a force F 2 i, then F 2 i = — F 12 . In
other words, to every action there is an equal and opposite reaction.
DEFINITIONS OF FORCE AND MASS
The concepts of force and mass used in the above axioms are as yet undefined, although
intuitively we have some idea of mass as a measure of the "quantity of matter in an object"
and force as a measure of the "push or pull on an object". We can however use the above
axioms to develop definitions [see Problem 2.28, page 49].
UNITS OF FORCE AND MASS
Standard units of mass are the gram (gm) in the cgs (centimetergramsecond) system,
kilogram (kg) in the mks (meterkilogramsecond) system and pound (lb) in the fps (foot
poundsecond) system. Standard units of force in these systems are the dyne, newton (nt)
and poundal (pdl) respectively. A dyne is that force which will give a 1 gm mass an accelera
tion of 1 cm/sec 2 . A newton is that force which will give a 1 kg mass an acceleration of
1 m/sec 2 . A poundal is that force which will give a 1 lb mass an acceleration of 1 ft/sec 2 .
For relationships among these units see Appendix A, page 341.
INERTIAL FRAMES OF REFERENCE. ABSOLUTE MOTION
It must be emphasized that Newton's laws are postulated under the assumption that
all measurements or observations are taken with respect to a coordinate system or frame
33
34
NEWTON'S LAWS OF MOTION. WORK, ENERGY AND MOMENTUM
[CHAP. 2
of reference which is fixed in space, i.e. is absolutely at rest. This is the socalled assump
tion that space or motion is absolute. It is quite clear, however, that a particle can be
at rest or in uniform motion in a straight line with respect to one frame of reference and
be traveling in a curve and accelerating with respect to another frame of reference.
We can show that if Newton's laws hold in one frame of reference they also hold in
any other frame of reference which is moving at constant velocity relative to it [see
Problem 2.3]. All such frames of reference are called inertial frames of reference or
Newtonian frames of reference. To all observers in such inertial systems the force
acting on a particle will be the same, i.e. it will be invariant. This is sometimes called
the classical principle of relativity.
The earth is not exactly an inertial system, but for many practical purposes can be
considered as one so long as motion takes place with speeds which are not too large. For
noninertial systems we use the methods of Chapter 6. For speeds comparable with the
speed of light (186,000 mi/sec), Newton's laws of mechanics must be replaced by Einstein's
laws of relativity or relativistic mechanics.
WORK
If a force F acting on a particle gives it a
displacement dr, then the work done by the
force on the particle is defined as
dW = Fdr
(3)
since only the component of F in the direction
of dr is effective in producing the motion.
The total work done by a force field (vector
field) F in moving the particle from point Pi
to point P 2 along the curve C of Fig. 21 is
given by the line integral [see Chap. 1, page 9].
Fig. 21
W =
Fdr = I Fdr = \ F
c *yPi "n
dr
(*)
where ri and r 2 are the position vectors of Pi and P 2 respectively.
POWER
The time rate of doing work on a particle is often called the instantaneous power, or
briefly the power, applied to the particle. Using the symbols W and <P for work and
power respectively we have .
r dt w
If F is the force acting on a particle and v is the velocity of the particle, then we have
<P = Fv (6)
KINETIC ENERGY
Suppose that the above particle has constant mass and that at times U and U it is
located at Pi and P 2 [Fig. 21] and moving with velocities vi = dn/dt and v 2 = drjdt
respectively. Then we can prove the following [see Problem 2.8].
CHAP. 2] NEWTON'S LAWS OF MOTION WORK, ENERGY AND MOMENTUM 35
Theorem 2 J. The total work done in moving the particle along C from Pi to P 2 is
given by
W = C Fdr = ^m(v\v\) (7)
If we call the quantity T = \mv^ (8)
the kinetic energy of the particle, then Theorem 2.1 is equivalent to the statement
Total Work done from Pi to P2 along C
= Kinetic energy at P2 — Kinetic energy at Pi
or, in symbols, W = T 2 Ti (10)
where Ti — ^mv 2 v T 2 — \mv\. ^
I
CONSERVATIVE FORCE FIELDS
Suppose there exists a scalar function V such that F =  vV. Then we can prove the
following [see Problem 2.15].
Theorem 2.2. The total work done in moving the particle along C from Pi to P 2 is
W = f p 2 Fdr = V(Pi)  V(P 2 ) (11)
In such case the work done is independent of the path C joining points Pi and P 2 . If the
work done by a force field in moving a particle from one point to another point is
independent of the path joining the points, then the force field is said to be conservative.
The following theorems are valid.
Theorem 23. A force field F is conservative if and only if there exists a continuously
differentiable scalar field V such that F=vF or, equivalently, if and only if
V X F = curl F = identically (12)
Theorem 2.4. A continuously differentiable force field F is conservative if and only
if for any closed nonintersecting curve C (simple closed curve)
i Fdr = (13)
i.e. the total work done in moving a particle around any closed path is zero.
POTENTIAL ENERGY OR POTENTIAL
The scalar V such that F=yF is called the potential energy, also called the scalar
potential or briefly the potential, of the particle in the conservative force field F. In such
case equation (11) of Theorem 2.2 can be written
Total Work done from Pi to P 2 along C
= Potential energy at Pi  Potential energy at P 2
or, in symbols, W = Vi  V 2 (15)
where Vi = V(Pi), V 2 = V(P 2 ).
36
NEWTON'S LAWS OF MOTIQN. WORK, ENERGY AND MOMENTUM [CHAP. 2
It should be noted that the potential is defined within an arbitrary additive constant.
We can express the potential as
V i=  f Fdr (16)
where we suppose that V = when r = r .
CONSERVATION OF ENERGY
For a conservative force field we have from equations (10) and (15),
T 2 Ti = Vi V a or T1 + V1 = T 2 + V 2
which can also be written \mv\ + Vi — \mv\ + V 2
(17)
(18)
The quantity E = T + V, which is the sum of the kinetic energy and potential energy, is
called the total energy. From (18) we s^ee that the total energy at Pi is the same as the
total energy at P 2 . We can state our results in the following
Theorem 2.5. In a conservative fofce field the total energy [i.e. the sum of kinetic
energy and potential energy] is a constant. In symbols, T + V = constant = E.
This theorem is often called the principle of conservation of energy.
IMPULSE
Suppose that in Fig. 21 the particle is located at Pi and P2 at times £1 and U where it
has velocities vi and v 2 respectively. Thb time integral of the force F given by
r
Ydt
(19)
is called the impulse of the force F. The following theorem can be proved [see Problem 2.18].
Theorem 2.6. The impulse is equal to the change in momentum; or, in symbols,
Xt 2
F
dt = m\ 2 — mvj
Pi
(20)
The theorem is true even when the mass is variable and the force is nonconservative.
TORQUE AND ANGULAR MOMENTUM
If a particle with position vector r moves in a
force field F [Fig. 22], we define
A = rxF
(21)
as the torque or moment of the force F about 0.
The magnitude of A is a measure of the j'turning
effect" produced on the particle by the fcjrce. We
can prove the following [see Problem 2.5J50]
Theorem 2.7.
rxF = 3r{m(rXv)}
(22)
Fig. 22
CHAP. 2] NEWTON'S LAWS OF MOTION. WORK, ENERGY AND MOMENTUM 37
The quantity
12 = m(rxv) = rxp (23)
is called the angular momentum or moment of momentum about 0. In words the theorem
states that the torque acting on a particle equals the time rate of change in its angular
momentum, i.e.,
 = w w
This theorem is true even if the mass m is variable or the force nonconservative.
CONSERVATION OF MOMENTUM
If we let F = in Newton's second law, we find
jT(rav) = or mv = constant (25)
This leads to the following
Theorem 2.8. If the net external force acting on a particle is zero, its momentum
will remain unchanged.
This theorem is often called the principle of conservation of momentum. For the case
of constant mass it is equivalent to Newton's first law.
CONSERVATION OF ANGULAR MOMENTUM
If we let A = in (2U) f we find
j
jr{m(rXv)} = or m(rXv) = constant (26)
This leads to the following
Theorem 2J9. If the net external torque acting on a particle is zero, the angular
momentum will remain unchanged.
This theorem is often called the principle of conservation of angular momentum.
NONCONSERVATIVE FORCES
If there is no scalar function V such that F =  yV [or, equivalently, if V x F *■ 0],
then F is called a nonconservative force field. The results (7), (20) and (2U) above hold
for all types of force fields, conservative or not. However, (11) and (17) or (18) hold only
for conservative force fields.
STATICS OR EQUILIBRIUM OF A PARTICLE
An important special case of motion of a particle occurs when the particle is, or appears
to be, at rest or in equilibrium with respect to an inertial coordinate system or frame of
reference. A necessary and sufficient condition for this is, from Newton's second law, that
F = (27)
i.e. the net (external) force acting on the particle be zero.
38 NEWTON'S LAWS OF MOTION. WORK, ENERGY AND MOMENTUM [CHAP. 2
If the force field is conservative with potential V, then a necessary and sufficient
condition for a particle to be in equilibrium at a point is that
F = . dV = dV = §V =
at the point. ' ^ Bx dy dz
STABILITY OF EQUILIBRIUM
If a particle which is displaced slightly from an equilibrium point P tends to return
to P, then we call P a point of stability or stable point and the equilibrium is said to be
stable. Otherwise we say that the point is one of instability and the equilibrium is
unstable. The following theorem is fundamental.
Theorem 2.10. A necessary and sufficient condition that an equilibrium point be one
of stability is that the potential V at the point be a minimum.
Solved Problems
NEWTON'S LAWS
2.1. Due to a force field, a particle of mass 5 units moves along a space curve whose
position vector is given as a function of time t by
r = (2t s + t)i + (St 4 t 2 + 8) j  12t 2 k
Find (a) the velocity, (b) the momentum, (c) the acceleration and (d) the force
field at any time t .
dr
(a) Velocity = v = ^ = (6* 2 + l)i + (12*3  2*)j  24tk
(6) Momentum = p = mv = 5v = (30*2 + 5)j + ( 60 f3 _ io*)j  120*
(c) Acceleration = a = ^ = ^  12ti + (36*2 _ 2 )j  24k
(d) Force = F = j^ = m^ = 60ti + (180t 2  10) j  120k
2.2. A particle of mass m moves in the xy plane so that its position vector is
r = a cos (d i + b sin <*>t j
where a, b and «> are positive constants and a> b. (a) Show that the particle moves
in an ellipse, (b) Show that the force acting on the particle is always directed
toward the origin.
(a) The position vector is
r = xi + y] = a cos at i + b sin at j
and so x = a cos at, y = b sin at which are
the parametric equations of an ellipse having
semimajor and semiminor axes of lengths a
and b respectively [see Fig. 23].
Since
(*/o) a + (y/b) 2 = cos 2 at + sin 2 at  1
the ellipse is also given by as 2 /a 2 + y 2 /b 2 = 1. Fig. 23
CHAP. 2]
NEWTON'S LAWS OF MOTION. WORK, ENERGY AND MOMENTUM
39
(6) Assuming the particle has constant mass m, the force acting on it is
d 2
„ dv d 2 r
F = m Tt = m w
m r£ [(a cos ut)i + (b sin «t)j]
= m[— w 2 a cos wt i — w 2 6 sin ut j]
— —mw 2 [a cos ut i + b sin ut j] = — mw 2 r
which shows that the force is always directed toward the origin.
2.3. Two observers O and 0', fixed relative to
two coordinate systems Oxyz and O'x'y'z'
respectively, observe the motion of a par
ticle P in space [see Fig. 24]. Show that
to both observers the particle appears to
have the same force acting on it if and
only if the coordinate systems are moving
at constant velocity relative to each other.
Let the position vectors of the particle in the
Oxyz and O'x'y'z' coordinate systems be r and r'
respectively and let the position vector of 0'
with respect to O be R = r — r\
Fig. 24
Relative to observers O and O' the forces acting on P according to Newton's laws are given
respectively by d 2 r < p r ,
F = m w' F ' = m W
The difference in observed forces is
F ~ F = m ^ (r " r) = m l#
and this will be zero if and only if
d 2 R
dt 2
=
dR
dt
= constant
i.e. the coordinate systems are moving at constant velocity relative to each other. Such coordinate
systems are called inertial coordinate systems.
The result is sometimes called the classical principle of relativity.
2.4. A particle of mass 2 moves in a force field depending on time t given by
F = 24t 2 i + (36£16)j  12tk
Assuming that at t = the particle is located at r = 3i  j + 4k and has velocity
vo = 6i + 15j  8k, find (a) the velocity and (b) the position at any time t.
(a) By Newton's second law,
2dv/dt = 24t 2 i + (36t16)j  12tk
or
dv/dt = 12tH + (18t8)j  6tk
Integrating with respect to t and calling c x the constant of integration, we have
v = UH + (9£ 2 8t)j  3t 2 k + c t
Since v = v = 6i + 15j  8k at t = 0, we have c x = 6i + 15j  8k and so
v = (4t 3 + 6)i + (9t 2 Bt+ 15)j  (3* 2 + 8)k
40 NEWTON'S LAWS OF MOTION. WORK, ENERGY AND MOMENTUM [CHAP. 2
(6) Since v = dr/dt, we have by part (a)
dr
^ = (4^3 + 6 )i + (9*2 _ 8t + 16) j _ (3t 2 + 8 ) k
Integrating with respect to t and calling c 2 the constant of integration,
r = (t 4 + 6«)i + (3t 3  4«2 + I5t)j  (tf* + 8t)k + c 2
Since r = r = 3i — j + 4k at t = 0, we have c 2 = 3i — j + 4k and so
r = (t* + 6t + 3)i + (3*3  4* 2 + 15t  l)j + (4  &  St)k
2.5. A constant force F acting on a particle of mass m changes the velocity from vi to V2
in time t.
(a) Prove that F = m(v 2 — vi)/t.
(b) Does the result in (a) hold if the force is variable? Explain.
(a) By Newton's second law, , , _
dv _ dv F /<v
mj = F or 77 = — (1)
dt dt m y '
Then if F and m are constants we have on integrating,
v = (F/m)t + c 2
At t = 0, v = v x so that c x = Vj i.e.
v = (F/m)t + vj (2)
At t = r, v = v 2 so that v 2 = (F/m)r + v t
i.e. F = m(v 2 — vi)/t (S)
Another method.
Write (1) as mdv = Fdt. Then since v=v! at t = and v = v 2 at t — r, we have
I md\ — \ Ydt or m(v 2 — v^ = Ft
vi
which yields the required result.
(b) No, the result does not hold in general if F is not a constant, since in such case we would not
obtain the result of integration achieved in (a).
2.6. Find the constant force in the (a) cgs system and (b) mks system needed to
accelerate a mass of 10,000 gm moving along a straight line from a speed of
54 km/hr to 108 km/hr in 5 minutes.
Assume the motion to be in the direction of the positive x axis. Then if v x and v 2 are the
velocities, we have from the given data V! = 54i km/hr, v 2 = 108i km/hr, m = 10,000 gm,
t = 5 min.
(a) In the cgs system
m = 10 4 gm, v x = 54i km/hr = 1.5 X lQ3i cm/sec, v 2 = 3.0 X 103i cm/sec, t = 300 sec
/v 2 — v t \ /1.5 X 10 3 i cm/sec \
Then F = ma = m ( . ) = (10 4 gm) )
V * / V 3 X 102 sec /
= 0.5 X 10 5 i gm cm/sec 2 = 5 X 10 4 i dynes
Thus the magnitude of the force is 50,000 dynes in the direction of the positive x axis.
(6) In the mks system
m = 10 kg, vj = 54i km/hr = 15i m/sec, v 2 = 30i m/sec, t = 300 sec
CHAP. 2] NEWTON'S LAWS OF MOTION. WORK, ENERGY AND MOMENTUM
41
Then
F = wa = m
v 2  Vl
= (10 kg)
15i m/sec
300 sec
= 0.5i kg m/sec 2 = 0.5i newtons
Thus the magnitude is 0.5 newtons in the positive x direction. This result could also have been
obtained from part (a) on noting that 1 newton = 10* dynes or 1 dyne = 10 newtons.
In this simple problem the unit vector i is sometimes omitted, it being understood that the
force F will have the direction of the positive * axis. However, it is good practice to work this
and similar problems with the unit vector present so as to emphasize the vector character of force,
velocity, etc This is especially important in cases where velocities may change their directions.
See, for example, Problem 2.46, page 56.
2.7. What constant force is needed to bring a 2000 lb mass moving at a speed of 60 mi/hr
to rest in 4 seconds?
We shall assume that the motion takes place in a straight line which we choose as the positive
direction of the x axis. Then using the English absolute system of units, we have
m = 2000 1b, v x = 60i mi/hr = 88i ft/sec, v 2 = Oi ft/sec, t = 4 sec
Then
F = wa = m
v 2  v x
'88i ft/sec \
= (2000 lb) ( )
4 sec /
= 4.4 X 10 4 i ft lb/sec 2 = 4.4 X 10 4 i poundals
Thus the force has magnitude 4.4 X 10* poundals in the negative x direction, i.e. in a direction
opposite to the motion. This is of course to be expected.
WORK, POWER, AND KINETIC ENERGY
2.8. A particle of constant mass m moves in space under the influence of a force field F
Assuming that at times U and U the velocity is vi and v 2 respectively, prove that
the work done is the change in kinetic energy, i.e.,
tt
X
Work done
 £•■*•  X.
F*\dt
•dv
= im f * d(v • v) = £mv 2 * = %mv\  Jmv 2
2.9. Find the work done in moving an object along a
vector r = 3i + 2j5k if the applied force is
F = 2i  j  k. Refer to Fig. 25.
Work done = (magnitude of force in direction
of motion) (distance moved)
= (Fcostf)(r) = Ft
= (2ijk)(3i + 2j5k)
= 62 + 5 = 9
r
Fig. 25
42 NEWTON'S LAWS OF MOTION. WORK, ENERGY AND MOMENTUM [CHAP. 2
2.10. Referring to Problem 2.2, (a) find the kinetic energy of the particle at points A
and B, (b) find the work done by the force field in moving the particle from A to B,
(c) illustrate the result of Problem 2.8 in this case and (d) show that the total work
done by the field in moving the particle once around the ellipse is zero.
(a) Velocity = v == dr/dt — — ua sin ut i + ub cos at j.
Kinetic energy = w 2 = ^m(u 2 a 2 sin 2 ut + u 2 b 2 cos 2 ut).
Kinetic energy at A [where cos ut — 1, sin ut = 0] = 4mw 2 6 2
Kinetic energy at B [where cos ut = 0, sin ut = 1] = ^mu 2 a 2
(6) Method 1. From part (6) of Problem 2.2,
= J F • dr = \ (—mu 2 r) • dr = — raw 2 I r • dr
J B B
d(r • r) = —^mu 2 r 2
A A
= \mu 2 a 2  mo> 2 6 2 = ^mw 2 (a 2  6 2 )
Work done
Method 2. We can assume that at A and B, t = and t = tt/2u respectively. Then:
Work done = I F • dr
J A
Xir/2u
[— mu 2 (a cos ut i + b sin ut j)] • [—ua sin ut i + ub cos ut j] dt
s>ir/2(>)
= I mw 3 (a 2 — 6 2 ) sin wt cos ut dt
o
ir/2w
= %mu 2 (a 2  b 2 ) sin 2 ut = %mu 2 (a 2  b 2 )
o
(c) From parts (a) and (6),
Work done = ww 2 (a 2  6 2 ) = \mu 2 a 2  \mu 2 b 2
— kinetic energy at A — kinetic energy at B
(d) Using Method 2 of part (6) we have, since t goes from to t = 2v/u for a complete circuit
around the ellipse,
27T/W
=
J..S7JVCU
mu z (a 2 — b 2 ) sin ut cos wt dt
o
= £mw 2 (a 2  6 2 ) sin 2 ut
Method 1 can also be used to show the same result.
2.11. Prove that if F is the force acting on a particle and v is the (instantaneous) velocity
of the particle, then the (instantaneous) power applied to the particle is given by
cp  fv
By definition the work done by a force F in giving a particle a displacement dr is
dW = F*dr
Then the (instantaneous) power is given by
as required.
* = ^ = '•! = '
CHAP. 2] NEWTON'S LAWS OF MOTION. WORK, ENERGY AND MOMENTUM 43
2.12. Find the (instantaneous) power applied to the particle in Problem 2.1 by the force
field.
By Problem 2.1, the velocity and force are given respectively by
v  (6* 2 + l)i + (12*3 _ 2t)j  24<k
F = 60ti + (180* 2  10)j  120k
Then the power [by Problem 2.11] is given by
q> = F • v = (60t)(6t 2 + 1) + (180t 2  10)(12*3  2t) + (120)(24t)
= 2160t 5  120* 3 + 2960£
2.13. Find the work done by the force in (a) Problem 2.6, (b) Problem 2.7.
(a) In the cgs system: v x = v x  = 1.5 X 10 3 cm/sec, v 2 = v 2  = 3.0 X 10 3 cm/sec, m = 10 4 gm.
Then by Problem 2.8,
Work done = change in kinetic energy
cm 2
= £(10 4 gm)(9.0 X 106 _ 2.25 X 106) _^
= 3.38 X lOio V£°*L = 3. 38xlo io/sn_n N ) (cm)
sec 2 \ sec z /
= 3.38 X 10 10 dyne cm = 3.38 X 10 10 ergs
In the mks system we have similarly:
Work done = £(10 kg)(900  225) ^
/ kg m \
= 3.38 X 10 3 ( j^g ) (m) = 3  38 x 10 newton meters
(6) As in part (a), ». 2
Work done = £(2000 lb)(88 2  2 ) i^
= 7.74 X 106(ft) (^f) = 7.74 X 10« ft pdl
CONSERVATIVE FORCE FIELDS, POTENTIAL ENERGY, AND
CONSERVATION OF ENERGY
2.14. Show that the force field F defined by
F = (y 2 z 8  6xz 2 )i + 2xyz 3 j + {Sxy 2 z 2  6x 2 z)k
is a conservative force field.
Method 1. The force field F is conservative if and only if curl F = V X F = 0. Now
V X F =
i j k
d/3x d/dy d/dz
y 2 z z _ q xz z 2xyz s 3xy 2 z 2  6x 2 z
= i \j{3xy 2 z 2 6x 2 z)  j^(2xyzS)
+ j \^(y 2 z^6xz 2 )  ^(SxyW  §x 2 z) 1
9a?
+ k ["^(fcsy* 8 )  j (y 2 z*  6xz 2 )~\
Then the force field is conservative.
44 NEWTON'S LAWS OF MOTION. WORK, ENERGY AND MOMENTUM [CHAP. 2
Method 2.
The force field F is conservative if and only if there exists a scalar function or potential
V(x,y, z) such that F = grad V = W Then
dx dy i dz K
= (y 2 z 3  6xz 2 )i + 2xyz 3 j + (3xy 2 z 2  6x 2 z)k
Hence if F is conservative we must be able to find V such that
dV/dx = 6xz 2  y 2 z 3 , dV/dy = 2xyz 3 , BV/dz = 6x 2 z  3xy 2 z 2 (1)
Integrate the first equation with respect to x keeping y and z constant. Then
V = 3x 2 z 2  xy 2 z 3 + g x {y, z) (2)
where g t (y, z) is a function of y and z.
Similarly integrating the second equation with respect to y (keeping x and z constant) and the
third equation with respect to z (keeping x and y constant), we have
V = xy 2 z 3 + g 2 (x, z) (3)
V = Zx 2 z 2 — xy 2 z 3 + g 3 (x, y) (4)
Equations (2), (3) and (4) yield a common V if we choose
ffi(V, *) = c, g 2 (x, z) = 3x 2 z 2 + c, g 3 (x, y)  c (5)
where c is any arbitrary constant, and it follows that
V = 3x 2 z 2 — xy 2 z 3 + c
is the required potential.
Method 3.
Fdr =  I (2/2*3 _ Qxz 2 )dx + 2xyz 3 dy + (3xy 2 z 2  6x 2 z)dz
.'o ^ (.x ,y ,z Q )
s*(x,y,z)
= I d(xy 2 z 3  3x 2 z 2 ) = 3x 2 z 2  xy 2 z 3 + c
(x ,y ,z )
where c = x y 2 z 3  3x 2 z^
2.15. Prove Theorem 2.2, page 35: If the force acting on a particle is given by F = — vV",
then the total work done in moving the particle along a curve C from Pi to P% is
W
We have
= f Frfr = V(Pi)  V(P 2 )
= J Fdr = J VV*dr = 1
W = \ Fdr = I VV*dr = I dV
Pi Pi J Pi
= V(P X )  V(P 2 )
Pi
2.16. Find the work done by the force field F of Problem 2.14 in moving a particle from
the point A(2,l,3) to 5(1,2,1).
Work done = f F'dr = f VV  dr
J A J A
J (1,2,1)
dV = V(x,y,z)
(2,1,3)
= — 3x 2 z 2 + xy 2 z 3 — c
(1,2,1)
(2,1,3)
(1,2,1)
= 155
(2,1,3)
CHAP. 2] NEWTON'S LAWS OF MOTION. WORK, ENERGY AND MOMENTUM 45
2.17. (a) Show that the force field of Problem 2.2 is conservative.
(6) Find the potential energy at points A and B of Fig. 23.
(c) Find the work done by the force in moving the particle from A to B and compare
with Problem 2.10(6).
(d) Find the total energy of the particle and show that it is constant, i.e. demonstrate
the principle of conservation of energy.
(a) From Problem 2(6), F = —ma 2 r = ma 2 {xi + yj). Then
i J k
V X F =
d/dx d/dy d/dz
—ma 2 x —mu 2 y
+ k \h { ~ mu2y) ~ ^ { ~ ma2x) \
=
Hence the field is conservative.
(6) Since the field is conservative there exists a potential V such that
W dV  dV u
Then dV/dx = ma 2 x, dV/dy = mw 2 y, dV/dz =
from which, omitting the constant, we have
V = \ma 2 x 2 + %ma 2 y 2 = $ma 2 (x 2 + y 2 ) = \mJh 2
which is the required potential.
(c) Potential at point A of Fig. 23 [where r = a] = $ma 2 a 2 .
Potential at point B of Fig. 23 [where r = b] = £mu 2 6 2 . Then
Work done from A to B = Potential at A — Potential at B
= \ma 2 a 2  £m<o 2 6 2 = $ma 2 (a 2  b 2 )
agreeing with Problem 2.10(6).
(d) By Problems 2.10(a) and part (6),
Kinetic energy at any point = T = %mv 2 = mf 2
= ra(<o 2 a 2 sin 2 at + a 2 b 2 cos 2 at)
Potential energy at any point = V = ^m« 2 r 2
= %ma 2 (a 2 cos 2 at + b 2 sin 2 at)
Thus at any point we have on adding and using sin 2 at + cos 2 at = 1,
T+V = $ma 2 (a 2 + b 2 )
which is a constant.
IMPULSE, TORQUE, ANGULAR MOMENTUM, AND
CONSERVATION OF MOMENTUM
2.18. Prove Theorem 2.6, page 36: The impulse of a force is equal to the change in
momentum.
By definition of impulse [see (19), page 36] and Newton's second law, we have
t2
C* ¥dt = J 2 ^( m v) dt = J d ^
(mv)
= mv 2 — mxi
46 NEWTON'S LAWS OF MOTION. WORK, ENERGY AND MOMENTUM [CHAP. 2
2.19. A mass of 5000 kg moves on a straight line from a speed of 540 km/hr to 720 km/hr
in 2 minutes. What is the impulse developed in this time?
Method 1.
Assume that the mass travels in the direction of the positive x axis. In the mks system,
_ _. n . km _ 540i X 1000 m 1Kvin2 . m
V, = 5401:; = = 1.5 X 10 2 1
hr 3600 sec sec
„™. km 720i X 1000 m nn „««. m
v 2 = 720i=^ = = 2.0Xl0 2 i —
hr 3600 sec sec
Then from Problem 2.18,
Impulse = m(v 2 — v x ) = (5000 kg)(0.5 X 10 2 i m/sec)
= 2.5 X 10 5 i kg m/sec = 2.5 X 10 5 i newton sec
since 1 newton = 1 kg m/sec 2 or 1 newton sec = 1 kg m/sec.
Thus the impulse has magnitude 2.5 X 10 5 newton sec in the positive x direction.
Method 2.
Using the cgs system, v x = 540i km/hr = 1.5 X 10 4 i cm/sec and v 2 = 720i km/hr =
2.0 X 10 4 i cm/sec. Then
Impulse = w(v 2  v t ) = (5000 X 10 3 gm)(0.5 X 10 4 i cm/sec)
= 2.50 X 10 10 igm cm/sec = 2.50 X 10 10 i dyne sec
since 1 dyne = 1 gm cm/sec 2 or 1 dyne sec = 1 gm cm/sec
Note that in finding the impulse we did not have to use the time 2 minutes as given in the
statement of the problem.
2.20. Prove Theorem 2.7, page 36: The moment of force or torque about the origin O of
a coordinate system is equal to the time rate of change of angular momentum.
The moment of force or torque about the origin O is
A = rXF = rXj (mv)
The angular momentum or moment of momentum about O is
B = m(r X v) = r X (mv)
Now we have 42L = ^(rXmv) = J^ X (mv) + rx(mv)
= v X (mv) + r X (mv) = + rXF = A
at
which gives the required result.
2.21. Determine (a) the torque and (6) the angular momentum about the origin for the
particle of Problem 2.4 at any time t.
(a) Torque A = rXF
= [(* 4 + 6* + 3)i + (3*3  4* 2 + 15*  1) j + (4  * 3  8*)k] X [24* 2 i + (36*  16)j  12*k]
i J k
t 4 + 6* + 3 3* 3  4* 2 + 15*  1 4  t 3  8*
24t 2 36*  16 12*
= (32*3 + 108*2 _ 260* + 64)i  (12* 5 + 192* 3  168* 2  36*) j
 (36* 5  80* 4 + 360*3 _ 240* 2  12* + 48)k
CHAP. 2] NEWTON'S LAWS OF MOTION. WORK, ENERGY AND MOMENTUM
47
(b) Angular momentum SI = r X (mv) = m(r X v)
= 2[(t* + 6* + 3)i + (3*3  4* 2 + 15*  l)j + (4  *3  8*)k]
X [(4*3 + 6)i + (9* 2  8* + 15)j  (3*2 + 8)k]
i J k
= 2 * 4 + 6* + 3 3* 3  4*2 + 15*  1 4  * 3  8*
4*3 + 6 9* 2  8* + 15 3* 2  8
= (8*4 + 36*3 _ 130*2 + 64t _ I04)i  (2*« + 48**  56* 3  18* 2  96)j
 (6*«  16*5 + 90*4 _ 80*3 _ 6*2 + 48*  102)k
Note that the torque is the derivative with respect to * of the angular momentum, illustrating
the theorem of Problem 2.20.
2.22. A particle moves in a force field given by F = r 2 r where r is the position vector of
the particle. Prove that the angular momentum of the particle is conserved.
The torque acting on the particle is
A = rXF = rX (r 2 r) = r 2 (r X r) =
Then by Theorem 2.9, page 37, the angular momentum is constant, i.e. the angular momentum is
conserved.
NONCONSERVATIVE FORCES
2.23. Show that the force field given by F = x 2 yzi  xyz^k is nonconservative.
We have i j k
V X F = d/dx d/dy BlBz
x 2 yz —xyz 2
Then since V X F # 0, the field is nonconservative.
= xz*\ + (x 2 y + yz 2 )}  x 2 zk
STATICS OF A PARTICLE
2.24. A particle P is acted upon by the forces Fi, F 2 , F 3 , F 4 , F 8 and F 6 shown in Fig. 26.
Represent geometrically the force needed to prevent P from moving.
Fig. 26
Fig. 27
The resultant R of the forces F v F 2 , F 3 , F 4 , F 5 and F 6 can be found by vector addition as
indicated in Fig. 27. We have R = F 1 + F 2 + F 3 F4 + F5 + F 6 . The force needed to prevent
P from moving is — R which is a vector equal in magnitude to R but opposite in direction and
sometimes called the equilibrant.
48
NEWTON'S LAWS OF MOTION. WORK, ENERGY AND MOMENTUM [CHAP. 2
2.25. A particle is acted upon by the forces Fi = 5i  lOj + 15k, F 2 = lOi + 25j  20k and
F 3 = 15i20j + 10k. Find the force needed to keep the particle in equilibrium.
The resultant of the forces is
R = F x + F 2 + F 3 = (5i  lOj + 15k) + (lOi + 25j  20k) + (151  20j + 10k)
= 30i  5j + 5k
Then the force needed to keep the particle in equilibrium is — R = — 30i + 5 j — 5k.
2.26. The coplanar forces as indicated in Fig. 28 act on a particle P. Find the resultant
of these forces (a) analytically and (b) graphically. What force is needed to keep
the particle in equilibrium?
Unit = 20 lb
V
''/
30 °/\
V
•3/
.#/
///30°
iV
Fig. 28 Fig. 29
(a) Analytically. From Fig. 28 we have,
F x = 160(cos45° i + sin 45° j), F 2 = 100( cos 30° i + sin 30° j),
F 3 = 120( cos 60° i  sin 60° j)
Then the resultant R is
R = F t + F 2 + F 3
= (160 cos 45°  100 cos 30°  120 cos 60°)i + (160 sin 45° + 100 sin 30°  120 sin 60°)j
= 33.46i + 59.21J
Writing R = R cos a i + R sin a j where a is the angle with the positive x axis measured
counterclockwise, we see that
R cos a  33.46, R sin a = 59.21
Thus the magnitude of R is R = V(33.46) 2 + (59.21) 2 = 68.0 lb, and the direction a with
the positive x axis is given by tan a = 59.21/(33.46) = 1.770 or a = 119° 28'.
(6) Graphically. Choosing a unit of 20 lb as shown in Fig. 29, we find that the resultant has
magnitude of about 68 lb and direction making an angle of about 61° with the negative
x axis [using a protractor] so that the angle with the positive x axis is about 119°.
A force — R, i.e. opposite in direction to R but with equal magnitude, is needed to keep P
in equilibrium.
STABILITY OF EQUILIBRIUM
2.27. A particle moves along the x axis in a force field having potential V = fax 2 , k > 0.
(a) Determine the points of equilibrium and (5) investigate the stability.
(a) Equilibrium points occur where VV = or in this case
dV/dx = kx = or x =
Thus there is only one equilibrium point, at x = 0.
CHAP. 2] NEWTON'S LAWS OF MOTION. WORK, ENERGY AND MOMENTUM
49
(6) Method 1.
Since d 2 V/dx 2 = k > 0, it follows that at x = 0, V is a minimum. Thus by Theorem 2.10,
page 38, * = is a point of stability. This is also seen from Problem 2.36 where it is shown
that the particle oscillates about x = 0.
V(x)
Method 2.
dV.
= — kxi. Then
We have F = VV = — ^i =
ax
when x > the particle undergoes a force to
the left, and when x < the particle under
goes a force to the right. Thus x = is a
point of stability.
Method 3.
The fact that x = is a minimum point can
be seen from a graph of V(x) vs x [Fig. 210].
Fig. 210
MISCELLANEOUS PROBLEMS
2.28. Show how Newton's laws can be used to develop definitions of force and mass.
Let us first consider some given particle P, assuming for the present that its mass ra P is not
defined but is simply some constant scalar quantity associated with P. Axiom 1 states that if P
moves with constant velocity (which may be zero) then the force acting on it is zero. Axiom 2
states that if the velocity is not constant then there is a force acting on P given by m P a P where
a P is the acceleration of P. Thus force is defined by axioms 1 and 2 [although axiom 1 is unnecessary
since it can in fact be deduced from axiom 2 by letting F = 0]. It should be noted that force is a
vector and thus has all the properties of vectors, in particular the parallelogram law for vector
addition.
To define the mass m P of particle P, let us now allow it to interact with some particular
particle which we shall consider to be a standard particle and which we take to have unit mass.
If a P and a s are the accelerations of particle P and the standard particle respectively, it follows
from axioms 2 and 3 that m P a P = — a s . Thus the mass m P can be defined as — a s /a P .
2.29. Find the work done in moving a particle once around a circle C in the xy plane, if
the circle has center at the origin and radius 3 and if the force field is given by
F = (2xy + z)i + (x + yz 2 )j + (Sx  2y + 4z)k
In the plane z = 0, F = (2x  y)\ + (x + y)j + (3x  2y)k and rfr = dx i + dy j so that the
work done is
f F'dr  f [(2xy)i + (x + y)j + (Sx2y)k]'[dxi + dyj]
J c J c
= I (2x — y) dx + (x + y) dy
J c
Choose the parametric equations of the circle as x = 3 cos t,
y = 3 sin t where t varies from to 2v [see Fig. 211]. Then the
line integral equals
,2tt
I [2(3 cos t)  3 sin t] [3 sin t] dt + [3 cos t + 3 sin t] [3 cos t] dt
X27T Q
(99 sin t cos t) dt = 9t   sin 2 1
= lSir
In traversing C we have chosen the counterclockwise direction indi
cated in Fig. 211. We call this the positive direction, or say that C
has been traversed in the positive sense. If C were traversed in the
clockwise (negative) direction the value of the integral would be — 18tt.
r = xi + yj
= 3 cos t i + 3 sin t j
Fig.2ll
50 NEWTON'S LAWS OF MOTION. WORK, ENERGY AND MOMENTUM [CHAP. 2
2.30. (a) If F = vV r , where V is singlevalued and has continuous partial derivatives,
show that the work done in moving a particle from one point Pi s (xi, yi, zi) in
this field to another point P 2 = (x 2 , 2/2, z 2 ) is independent of the path joining the
two points.
(b) Conversely, if J F'dr is independent of the path C joining any two points,
show that there exists a function V such that F=vV.
ft pP 2
(a) Work done = I F«dr = — I VV»dr
J *x J P!
" J Pi (to 1 + aP + ^kjidxi + dyj + dzk)
J Pl d % ty dz
= f 2<iV = V(P 1 )V(P 2 ) = V(x x ,y x ,z x )  V(x 2 ,y 2 ,z 2 )
Then the integral depends only on points P x and P 2 and not on the path joining them.
This is true of course only if V(x,y,z) is singlevalued at all points P x and P 2 .
(6) Let F = F x \ + F 2 j + F 3 k. By hypothesis, I F • dx is independent of the path C joining any
two points, which we take as (x x ,y x , z x ) and (x,y,z) respectively. Then
J % (x,y,z) p(.x,y,z)
F • dt =  I (F x dx + F 2 dy + F s dz)
(.x i ,y 1 ,z 1 '> J <.x v y lt zi>
is independent of the path joining (x x , y x , z x ) and (x,y,z). Thus
V(x, y,z) =  I [F x (x, y, z) dx + F 2 (x, y, z) dy + F 3 (x, y, z) dz]
where C is a path joining (x x ,y x ,z x ) and (x,y,z). Let us choose as a particular path the
straight line segments from (x it y v z x ) to (x, y x , z x ) to (x, y, z x ) to (x, y, z) and call V(x, y, z) the
work done along this particular path. Then
XX pV pz
F x (x,y x ,z x )dx  J F 2 (x,y,z x )dy  J F 3 (x,y,z)dz
"i *i ^1
It follows that
— = F 3 {x,y,z)
dV C dF 3 C z 9F 2
7 = F2(x,y,z x )  I — — (x,y,z)dz = F 2 {x,y,z x )  I — (x,y,z)dz
dy ^ Xl dy *s Zl dz
z
= F 2 (x,y,z x )  F 2 (x,y,z)
*i
= F 2 (x,y,z x )  F 2 (x,y,z) + F 2 (x,y,z x ) = F 2 (x,y,z)
§V_
dx
C v SF 2 f 8F a
= f x (x, y x , z x )  j ^(*> v> *i) d y  j fate* y> *) dz
J'* dF x C* dF x
r{x,y,z x )dy  I — (x,y,z) dz
j/j <>y ^ zx °Z
y \z
= F x (x,y x ,z x )  F x (x,y,z x )  F x (x,y,z)\
v\ \*i
= F x (x,y x ,z x )  F x (x,y,z x ) + F(x,y x ,z x )  F x (x,y,z) + F(x,y,z x ) = F x (x,y,z)
CHAP. 2] NEWTON'S LAWS OF MOTION. WORK, ENERGY AND MOMENTUM
51
Then
F = r.l+J.J+F.k = fififk = W
3»" dy' dz
Thus a necessary and sufficient condition that a field F be conservative is that curl F = V X F = 0.
2.31. (a) Show that F = (2xy + z*)i + x 2 j + Sxz 2 k is a conservative force field, (b) Find
the potential, (c) Find the work done in moving an object in this field from (1, 2, 1)
to (3,1,4).
(a) A necessary and sufficient condition that a force will be conservative is that curl F = V X F — 0.
= 0. Thus F is a conservative force field.
Now V X F =
i J k
d/dx d/dy d/dz
2xy + z* x 2 Zxz 2
(6) As in Problem 2.14, Methods 2 or 3, we find V = ~(x 2 y + xz 3 ).
(3,1,4)
(c) Work done = — (x 2 y + xz 3 )
= 202.
(1,2,1)
F'dr is independent of the path joining any two points Pi and P 2
Pi r
in a given region, then i F • dt = f or all closed paths in the region and conversely.
Let P 1 AP 2 BP 1 [see Fig. 212] be a closed curve. Then ^^^
£ F*dr = J F • dr = J F • dr + J F • dr ^^^^^^ \p
J PiAPtBP! PiAP 2 P2BP1 S^ \ 2
= J Fdr J Fdr =
PiAP 2 PiBPa
since the integral from P x to P 2 along a path through A is
the same as that along a path through B, by hypothesis.
L^
Conversely if <i> F*dr = 0, then
J F'dr = J F'dr + J F'dr = J F»dr  J
P^PiBP! PiAP 2
, $ Fdr = / F
A
Fig. 212
F • dr =
P 2 BP t
P1AP2
P 1 BP 2
so that, I F»dr = I F«dr
PiAP 2 PiBP 2
2.33. (a) Show that a necessary and sufficient condition that Fi dx + F 2 dy + Fs dz be an
exact differential is that VxF = where F = Fii + F 2 j+F 3 k.
(b) Show that (y 2 z? cos x  4x s z) dx + 2z s y sin x dy + (Sy 2 z 2 sin x  x 4 ) dz is an exact
differential of a function <f> and find <f>.
(a) Suppose F x dx + F 2 dy + F 3 dz = d<f> = ^dx + ^dy + ^dz, an exact differential. Then
since x, y and z are independent variables,
52 NEWTON'S LAWS OF MOTION. WORK, ENERGY AND MOMENTUM [CHAP. 2
30 jt, _ d<f> „ _ 30
dx dy dz
and so F = F x i + F 2 j + F s k = *i + * j + ^k = V*. Thus VXF = VXV^ = 0.
dx dy dz ^
Conversely if V X F = 0, then F = V> and so F'dr = V#*dr = d<f>, i.e. Fj da; +
F 2 dy + F 3 dz = d<p, an exact differential.
(b) F = (y 2 z s cos a; — Ax z z) i + 2z 3 j/ sin x j + (32/ 2 « 2 sin x — x*) k and V X F is computed to be
zero, so that by part (a) the required result follows.
2.34. Referring to Problem 2.4 find (a) the kinetic energy of the particle at t  1 and
£ = 2, (6) the work done by the field in moving the particle from the point where
t = 1 to the point where t = 2, (c) the momentum of the particle at t = 1 and t = 2
and (d) the impulse in moving the particle from t = 1 to £ = 2.
(a) From part (a) of Problem 2.4,
v = (4*3 + 6)i + (9i 2  8t + 15)j  (St 2 + 8)k
Then the velocities at t = 1 and t = 2 are
v x = lOi + 16j  Ilk, v 2 = 38i + 35j  20k
and the kinetic energies at t = 1 and t = 2 are
T x = £ mv ? = £(2)[(10)2 + (16)2 + (11)2] = 477, T 2 = £mv 2 = 3069
(6) Work done = f F • dr
s.
X
[24* 2 i + (36t  16)j  12tk] • [(4« 3 + 6)i + (9* 2  8t + 15)j  (3t 2 + 8)k]dt
2
[(24t 2 K4*3 + 6) 4 (36£  16)(9t 2 8t + 15) + (12t)(3« 2 + g)]dt = 2592
Note that by part (a) this is the same as the difference or change in kinetic energies
3069477 = 2592, illustrating Theorem 2.1, page 35, that Work done = change in kinetic
energy.
(c) By part (a) the momentum at any time t is
p = mv = 2v = (8« 3 + 12)i + (18t 2  16* + 30)j  (6t 2 + 16)k
Then the momenta at t = 1 and t = 2 are
Pi = 20i + 32j  22k, p 2 = 76i + 70j  40k
(d) Impulse = ) F dt
J t=i
.2
[24t*i + (36t  16) j  12tk]dt = 56i + 38 j  18k
't=i
Note that by part (6) this is the same as the difference or change in momentum, i.e.
P2 ~ Pi = ( 76i + 7 °J ~ 40k )  ( 20i + 32 J ~~ 22k > = 56i + 38 ' ~ 18k ' illu strating Theorem 2.6,
page 36, that Impulse = change in momentum.
=/,:
2.35. A particle of mass m moves along the x axis under the influence of a conservative
force field having potential V(x). If the particle is located at positions x x and x 2 at
respective times U and U, prove that if E is the total energy,
♦ *  [™ C H dx
* 2 " *» " \ 2 X t y/E  V(x)
CHAP. 2] NEWTON'S LAWS OF MOTION. WORK, ENERGY AND MOMENTUM 53
By the conservation of energy,
Kinetic energy + Potential energy = E
$m{dx/dt) 2 + V{x) = E
Then {dx/dt) 2 = (2/m){E  V(x)} (1)
from which we obtain on considering the positive square root,
dt = yfm/2(dx/\/EV{x))
Hence by integration,
J ti dt ~ ****  \f f Xi Ve^vW)
2.36. (a) If the particle of Problem 2.35 has potential V = fax 2 and starts from rest at
x = a, prove that x = acosy^Jmt and (b) describe the motion.
(a) From (i) of Problem 2.35, (dx/dt) 2 = (2/m)(E  fax 2 ). Since dx/dt = where x = a, we
find E  faa 2 so that
(dx/dt) 2 = (K/m)(a 2 x 2 ) or dx/y/a 2  x 2 = ±y/Umdt
Integration yields sin 1 (x/a) = ± y/ic/m t + c x . Since x = a at t = 0, c l = v/2. Then
sin _1 (*/a) = ± sjic/m t + jt/2 or * = a sin (jt/2 ± y/ic/m t) = a co^yfic/mt
(6) The particle oscillates back and forth along the x axis from x = a to x — —a. The time for
one complete vibration or oscillation from x — a back to x = a again is called the period of the
oscillation and is given by P = 2v VWk.
2.37. A particle of mass 3 units moves in the xy plane under the influence of a force field
having potential V = l2x{Zy  4x). The particle starts at time t = from rest at
the point with position vector lOi  lOj. (a) Set up the differential equations and
conditions describing the motion. (6) Solve the equations in (a), (c) Find the
position at any time, (d) Find the velocity at any time.
(a) Since V = 12x(3y  4*) = S6xy  48a 2 , the force field is
F = ~ Vy = f^^f 1 ' = (362/ + 96*)i36*j
Then by Newton's second law,
d 2 *
3 ;/72 ~ (~36y + 96x)i  36;rj
dt 2
or in component form, using r = xi + yj,
dtx/dt 2 = 12y + 32a, cPy/dt 2 = 12* (1)
where x = 10, x = 0, y = 10, y = at t = (2)
using the fact that the particle starts at r = lOi — lOj with velocity v = f = 0.
(6) From the second equation of (i), * = —^d 2 yfdt 2 . Substitution into the first equation of (1)
yields
d*y/dt*  32 d 2 y/dt 2  144y = (3)
If a is constant then y = e«* is a solution of ($) provided that
«4  32a 2  144 = 0, i.e. (a 2 + 4)(a 2  36) = or a = ±2i, a = ±6
54 NEWTON'S LAWS OF MOTION. WORK, ENERGY AND MOMENTUM [CHAP. 2
Thus solutions are e 2it , e~ 2it , e 6t , e~ 6t or cos2t, sin2t, e 6t , e 6t [in terms of real functions]
and the general solution is
y = c t cos2t + c 2 sin2* + c 3 e 6t + c 4 e~ 6t (4)
Thus from a; = ~^cPy/dt 2 we find, using (.4),
a? = ^ Cos 2t + \c 2 sin 2*  3c 3 e«  3c 4 e~ 6t (5)
Using the conditions (2) in (4) and (5), we obtain
£<?!  3c 3  3c 4 = 10, §c 2  18c 3 + 18c 4 = 0,
c i + c 3 + c 4 = — 10, 2c 2 + 6c 3 — 6c 4 =
Solving simultaneously, c x = —6, c 2 = 0, c 3 = —2, c 4 = —2 so that
a? = 6 cos2t 2e 6t  2e~ 6 *, y = 2 cos 2t + 6e 6t + 6e«
(c) The position at any time is
r = xi + yj = (6 cos2t  2e 6t  2e~ flt )i + (2 cos 2* + 6e 6t + 6e~ 6t )j
(d) The velocity at any time is
v = r = xi + yj = (12 sin 2*  12e«' + 12e" 6t )i + (4 sin 2* + 36e 6 '  36e~ 6t )j
In terms of the hyperbolic functions
sinh at = £(««* ««*)» cosh at = ^(e at \ e~ at )
we can also write
r = (6 cos 2t  4 cosh 6t)i + (2 cos 2t + 12 cosh 6t)j
v = r = (12 sin 2t  24 sinh 6t)i + (4 sin It + 72 sinh 6«)j
2.38. Prove that in polar coordinates (r, 0),
Let VV = Gri + H*! (i)
where G and H are to be determined. Since dr = dx i + dy j we have on using a? = r cos ,
y = r sine and Problem 1.47(6), page 25,
dr = (cos * dr — r sin *)(cos $ r x — sin 9^ + (sin tf dr + r cos d*)(sin e r t + cos # x )
or dr = dr r x + r dff # x (2)
Now VV'dr = dV = ^dr + ^de
Using (i) and (2) this becomes
(Gr x + H9 1 )(drr 1 + rd69i) = Gdr + Hrde = jp dr + 'gf de
so that G ~ Ifr' H ~ rle
Then (1) becomes VV = ^T r i + ^ a7 # i
2.39. According to the theory of relativity, the mass m of a particle is given by
— m ° mp
W ~ VI  vVc 2 VI )8 2
where v is the speed, m the rest mass, c the speed of light and p = v/c.
CHAP. 2] NEWTON'S LAWS OF MOTION. WORK, ENERGY AND MOMENTUM 55
(a) Show that the time rate of doing work is given by
moc 2 ^!/? 2 )" 2
(b) Deduce from (a) that the kinetic energy is
T = (mmo)c 2 = m c 2 {(l/3 2 ) 1/2 l}
(c) If v is much less than c, show that T = %mv 2 approximately.
(a) By Newton's second law, , , / m v
F = jT(mv) =
dt dt \y/±
Then if W is the work done,
dW _ d / "W \ d f P \ 2 d / 1 \
as proved by direct differentiation.
(6) Since Work done = change in kinetic energy, we have
Time rate of doing work = time rate of change in kinetic energy
u _* / x dW dT 2 d / 1 \
or by part (a), ~dT = Hi = m « c *dt (^f^p)
WoC 2
Integrating, T = + c x
To determine Cj note that, by definition, T = when v = or /? = 0, so that Cj = — m «
Hence we have, as required,
T = ——z^^ — m c 2 = (m — mo)c 2
(c) For /? < 1 we have by the binomial theorem,
1 = (I/?*)"* = 1 + 1^2 + 1^1^ + 1^111^+...
r i v 2 i i
Then T = m c 2 1 +  ^ + • • • — m c 2 = ^mv 2 approximately
$ Supplementary Problems
NEWTON'S LAWS
2.40. A particle of mass 2 units moves along the space curve defined by r = (4t 2 — t 3 )i — 5tj + (t* — 2)k.
Find (a) the momentum and (6) the force acting on it at t = 1.
Ans. (a) 10i10j + 8k, (6) 4i + 24k
2.41. A particle moving in a force field F has its momentum given at any time t by
p = 3e _t i — 2 cos t j — 3 sin t k
Find F. Ans. — 3e~*i + 2 sin £ j — 3 cos t k
2.42. Under the influence of a force field a particle of mass m moves along the ellipse
r = a cos at i + b sin ut j
If p is the momentum, prove that (o)rXp = mabak, (6) r • p = £m(& 2 — a 2 ) sin 2at.
56 NEWTON'S LAWS OF MOTION. WORK, ENERGY AND MOMENTUM [CHAP. 2
2.43. If F is the force acting on the particle of Problem 2.42, prove that r X F = 0. Explain what this
means physically.
2.44. A force of 100 dynes in the direction of the positive x axis acts on a particle of mass 2 gm for
10 minutes. What velocity does the particle acquire assuming that it starts from rest?
Arts. 3 X 10 4 cm/sec
2.45. Work Problem 2.44 if the force is 20 newtons and the mass is 10 kg. Ans. 1200 m/sec
2.46. (a) Find the constant force needed to accelerate a mass of 40 kg from the velocity 4i — 5j + 3k
m/sec to 8i + 3j — 5k m/sec in 20 seconds. (6) What is the magnitude of the force in (a)?
Ans. (a) 8i + 16j  16k newtons or (8i + 16j  16k) X 10 5 dynes
(6) 24 newtons or 24 X 10 5 dynes
2.47. An elevator moves from the top floor of a tall building to the ground floor without stopping.
(a) Explain why a blindfolded person in the elevator may believe that the elevator is not moving
at all. (b) Can the person tell when the motion begins or stops? Explain.
2.48. A particle of unit mass moves in a force field given in terms of time t by
F = (6«8)i  60t3j + (20t3 + 36«2)k
Its initial position and velocity are given respectively by r = 2i — 3k and v = 5i + 4j. Find the
(a) position and (b) velocity of the particle at t = 2.
Ans. (a) 4i  88j + 77k, (6) i  236j + 176k
2.49. The force acting on a particle of mass m is given in terms of time t by
F = a cos ut i + b sin at j
If the particle is initially at rest at the origin, find its (a) position and (6) velocity at any later time.
Am. (a) — ^(1 — cos at) i H ^(at — sin at) j, (6) — sin at i H (1 — cos at) j
ma 2 ma 2 ma ma
WORK, POWER AND KINETIC ENERGY
2.50. A particle is moved by a force F = 20i — 30j + 15k along a straight line from point A to point B
with position vectors 2i + 7j — 3k and 5i — 3 j — 6k respectively. Find the work done.
Ans. 315
2.51. Find the kinetic energy of a particle of mass 20 moving with velocity 3i — 5j + 4k. Ans. 500
2.52. Due to a force field F, a particle of mass 4 moves along the space curve r = (3t 2 — 2t)\ + t 3 j — t 4 k.
Find the work done by the field in moving the particle from the point where t = 1 to the point
where t = 2. Ans. 2454
2.53. At one particular instant of time a particle of mass 10 is traveling along a space curve with velocity
given by 4i + 16k. At a later instant of time its velocity is 8i — 20j. Find the work done on the
particle between the two instants of time. Ans. 192
2.54. Verify Theorem 2.1, page 35 for the particle of Problem 2.52.
2.55. A particle of mass m moves under the influence of the force field given by F = a(sin at i + cos at j).
If the particle is initially at rest at the origin, prove that the work done on the particle up to time t
is given by (a 2 /m<o 2 )(l — cos at).
2.56. Prove that the instantaneous power applied to the particle in Problem 2.55 is (a 2 /mw) sin ut.
2.57. A particle moves with velocity 5i3j + 6k under the influence of a constant force F = 20i +
10j + 15k. What is the instantaneous power applied to the particle? Ans. 160
CHAP. 2] NEWTON'S LAWS OF MOTION. WORK, ENERGY AND MOMENTUM 57
CONSERVATIVE FORCE FIELDS, POTENTIAL ENERGY AND
CONSERVATION OF ENERGY
2.58. (a) Prove that the force field F = (y 2 — 2xyz s )i + (3 + 2xy — x 2 z 3 )j + (6z 3  3x 2 yz 2 )k is conservative.
(6) Find the potential V associated with the force field in (a).
Ans. (b) xy 2 — x 2 yz 3 + 3y + § z 4
2.59. A particle moves in the force field of Problem 2.58 from the point (2, — 1, 2) to (—1, 3, —2). Find the
work done. Ans. 55
2.60. (a) Find constants a, b, c so that the force field defined by
F = (x + 2y + az)i + (bx 3y z)j + (4x + cy + 2z)k
is conservative.
(6) What is the potential associated with the force field in (a)?
Ans. (a) a = 4, 6 = 2, c = 1 (6) V = \x 2 + f y 2  z 2  2xy  4xz + yz
2.61. Find the work done in moving a particle from the point (1,1,2) to (2,3,1) in a force field with
potential V = x 3 — y 3 + 2xy — y 2 + 4x. Ans. 15
2.62. Determine whether the force field F = (x 2 y — z 3 )i + (3xyz + xz 2 )j + (2x 2 yz + j/z 4 )k is conservative.
Ans. Not conservative
2.63. Find the work done in moving a particle in the force field F = 3x 2 i + (2xz — y)j + zk along
(a) the straight line from (0, 0, 0) to (2, 1, 3), (6) the space curve x = 2t 2 , y = t, z = 4t 2 —t from t =
to t = l. Is the work independent of the path? Explain. Ans. (a) 16, (6) 14.2
2.64. (a) Evaluate d> F • dt where F = (x — 3y)i + (y — 2#)j and C is the closed curve in the xy plane
J c
x — 2 cos t, y = 3 sin t from t = to t = 2v. (b) Give a physical interpretation to the result in (a).
Ans. (a) 6jt if C is traversed in the positive (counterclockwise) direction.
2.65. (a) Show that the force field F = — kt^t is conservative.
(6) Write the potential energy of a particle moving in the force field of (a).
(c) If a particle at mass m moves with velocity v = dr/dt in this field, show that if E is the constant
total energy then ^m(dr/dt) z + ^kt 5 = E. What important physical principle does this illustrate?
2.66. A particle of mass 4 moves in the force field defined by F = — 200r/r3. (a) Show that the field is
conservative and find the potential energy. (6) If a particle starts at r = 1 with speed 20, what will
be its speed at r = 2? Ans. (a) V  200/r, (6) 15^
IMPULSE, TORQUE AND ANGULAR MOMENTUM.
CONSERVATION OF MOMENTUM
2.67. A particle of unit mass moves in a force field given by F = (3t 2  te)i + (12t  6)j + (6*  12^
where t is the time, (a) Find the change in momentum of the particle from time t — 1 to t = 2.
(6) If the velocity at * = 1 is 4i — 5j + 10k, what is the velocity at * = 2?
Ans. (a) i + 12j  19k, (6) 5i + 7j  9k
2.68. A particle of mass m moves along a space curve defined by r = a cos at i + 6 sin at j. Find
(a) the torque and (b) the angular momentum about the origin. Ans. (a) 0, (6) 2mabak
2.69. A particle moves in a force field given by F = <f>(r) r. Prove that the angular momentum of the
particle about the origin is constant.
2.70 Find (a) the torque and (6) the angular momentum about the origin at the time t = 2 for the
particle of Problem 2.67, assuming that at t = it is located at the origin.
Ans. (a)  (36i + 128j + 60k), (6)  44i + 52j + 16k
2.71. Find the impulse developed by a force given by F = 4£i + (6t 2 — 2)j + 12k from t = to t = 2.
Ans. 8i + 12j + 24tk
58
NEWTON'S LAWS OF MOTION. WORK, ENERGY AND MOMENTUM [CHAP. 2
2.72. What is the magnitude of the impulse developed by a mass of 200 gm which changes its velocity from
5i — 3j + 7k m/sec to 2i + 3j + k m/sec? Ans. 1.8 X 10 5 dyne sec or 1.8 newton sec
STATICS OF A PARTICLE
2.73. A particle is acted upon by the forces F x = 2i + aj  3k, F 2 = 5i + cj + 6k, F 3 = 6i  5j + 7k,
F 4 = ci — 6j + ak. Find the values of the constants a, b, c in order that the particle will be in
equilibrium. Ans. a — 1, 6 = 11, c = 4
2.74. Find (a) graphically and (6) analytically the result
ant force acting on the mass m of Fig. 213 where
all forces are in a plane.
Ans. (6) 19.5 dynes in a direction making an angle
85°22' with the negative x axis
2.75. The potential of a particle moving in the xy plane is
given by V = 2x 2 — 5xy + Sy 2 + 6x7y. (a) Prove
that there will be one and only one point at which a
particle will remain in equilibrium and (6) find the
coordinates of this point. Ans. (6) (1, 2)
2.76. Prove that a particle which moves in a force field
of potential
V  x 2 + Ay 2 + z 2  Axy  4yz + 2xz  4a; + Sy  4z
can remain in equilibrium at infinitely many points
and locate these points.
Ans. All points on the plane x — 2yVz — 2
Fig. 213
STABILITY OF EQUILIBRIUM
2.77. A particle moves on the x axis in a force field having potential V = x 2 (6 — x).
(a) Find the points at equilibrium and (6) investigate their stability.
Ans. x = is a point of stable equilibrium; x = 4 is a point of unstable equilibrium
2.78. Work Problem 2.77 if (a) V = a; 4  8a; 3  6a; 2 + 24a;, (6) V = x 4 .
Ans. (a) x\,2 are points of stable equilibrium; x — 1 is a point of unstable equilibrium.
(6) x — is a point of stable equilibrium
2.79. Work Problem 2.77 if V = sin 2^a;.
Ans. If n  0, ±1, ±2, ±3, . . . then x = f + n are points of stable equilibrium, while x = J + n
are points of unstable equilibrium.
2.80. A particle moves in a force field with potential V = x 2 + y 2 + z 2  Sx + 16y  4z. Find the points
of stable equilibrium. Ans. (4, —8, 2)
MISCELLANEOUS PROBLEMS
2.81. (a) Prove that F = {y 2 cos x + z s )i + (2y sin x  4)j + (3a;z 2 + 2)k
is a conservative force field. (6) Find the potential corresponding
to F. (c) Find the work done in moving a particle in this field from
(0,1,1) to (tr/2, 1,2).
Ans. (a) V = y 2 sin x + xz 3  4y + 2z + c, (6) 15 + 4n
2.82. A particle P is acted upon by 3 coplanar forces as indicated in
Fig. 214. Find the force needed to prevent P from moving.
Ans. 323 lb in a direction opposite to 150 lb force
2.83. (a) Prove that F = r^x is conservative and (6) find the correspond
ing potential. Ans. (6) V = ^r 4 + c
100 lb
Fig. 214
CHAP. 2] NEWTON'S LAWS OF MOTION. WORK, ENERGY AND MOMENTUM 59
2.84. Explain the following paradox: According to Newton's third law a trailer pulls back on an auto
mobile to which it is attached with as much force as the auto pulls forward on the trailer. Therefore
the auto cannot move.
2.85. Find the potential of a particle placed in a force field given by F = — Kr~ n r where k and n are
constants. Treat all cases.
2.86. A waterfall 500 ft high has 440,000 ft 3 of water flowing over it per second. Assuming that the
density of water is 62.5 lb/ft 3 and that 1 horsepower is 550 ft lb/sec, find the horsepower of the
waterfall. Ans. 25 X 10 6 hp
2.87. The power applied to a particle by a force field is given as a function of time t by <P(t) = 3* 2 — At + 2.
Find the work done in moving the particle from the point where t = 2 to the point where t = 4.
Ans. 36
2.88. Can the torque on a particle be zero without the force being zero? Explain.
2.89. Can the force on a particle be zero without the angular momentum being zero? Explain.
2.90. Under the influence of a force field F a particle of mass 2 moves along the space curve
r = 6t 4 i — 3* 2 j + (4t 3 — 5)k. Find (a) the work done in moving the particle from the point where
t — to the point where t = l, (6) the power applied to the particle at any time.
Ans. (a) 756 (6) 72*(48i 4 + St + 1)
2.91. A force field moves a particle of mass m along the space curve r = acos«£i+ ftsinwi j. (a) What
power is required? (6) Discuss physically the case a — b. Ans. (a) m(a 2 — 6 2 )« 3 sinwt coswi
2.92. The angular momentum of a particle is given as a function of time t by
12 = 6tH  (2t + l)j + (12i 3  8* 2 )k
Find the torque at the time t = l. Ans. 12i — 2j + 20k
2.93. Find the constant force needed to give an object of mass 36,000 lb a speed of 10 mi/hr in 5 minutes
starting from rest. Ans. 1760 poundals
2.94. A constant force of 100 newtons is applied for 2 minutes to a 20 kg mass which is initially at rest,
(a) What is the speed achieved? (6) What is the distance traveled?
Ans. (a) 600 m/sec, (6) 36,000 m
2.95. A particle of mass m moves on the x axis under the influence of a force of attraction toward origin O
given by F = — (k/x 2 )L If the particle starts from rest at x = a, prove that it will arrive at O in a
time given by ^iray 'ma/2*.
2.96. Work Problem 2.95 if F = ( K /x*)i.
2.97. A particle of mass 2 units moves in the force field F = t 2 i — Btj + (t + 2)k where t is the time.
(a) How far does the particle move from * = to t = 3 if it is initially at rest at the origin?
(6) Find the kinetic energy at times t = 1 and t = 3. (c) What is the work done on the particle by
the field from t = 1 to t = 3? (d) What is the power applied to the particle at * = 1? (e) What is the
impulse supplied to the particle at t = 1 ?
2.98. At f = 0a particle of unit mass is at rest at the origin. If it is acted upon by a force F = 100te _2t i,
find (a) the change in momentum of the particle in going from time t = 1 to t = 2, (6) the velocity
after a long time has elapsed. Ans. (a) 25e _2 (3 — 5e 2 )i, (6) 25
60 NEWTON'S LAWS OF MOTION. WORK, ENERGY AND MOMENTUM [CHAP. 2
2.99. A particle of mass 3 units moves in the xy plane under the influence of a force field having potential
V = 6» 3 + 12y 3 + S6xy — 48s 2 . Investigate the motion of the particle if it is displaced slightly
from its equilibrium position.
[Hint. Near x = 0, y = the potential is very nearly 36xy  48x 2 since 6a; 3 and 12y s are negligible.]
2.100. A particle of unit mass moves on the x axis under the influence of a force field having potential
V = 6x(x — 2). (a) Show that x = 1 is a position of stable equilibrium. (6) Prove that if the mass
is displaced slightly from its position of equilibrium it will oscillate about it with period equal to
4rv / 3.
[Hint. Let x = 1 + u and neglect terms in u of degree higher than one.]
2.101. A particle of mass m moves in a force field F = — kx\. (a) How much work is done in moving the
particle from x = x x to x = x 2 ' (*») If a unit particle starts at x = x lt wit h speed v lt what is its speed
on reaching x = x 2 1 Ans. (a) %k{x\  x\), (6) V v 2 + (ic/m)(cc 2  a; 2 )
2.102. A particle, of mass 2 moves in the xy plane under the influence of a force field having potential
y = z 2 + y 2. The particle starts at time t = from rest at the point (2, 1). (a) Set up the differential
equations and conditions describing the motion. (6) Find the position at any time t. (c) Find the
velocity at any time t.
2.103. Work Problem 2.102 if V  Bxy.
2.104. Does Theorem 2.7, page 36, hold relative to a noninertial frame of reference or coordinate system?
Prove your answer.
2.105. (a) Prove that if a particle moves in the xy plane under the influence of a force field having potential
V = 12x(Sy  4x), then x = 0, y = is a point of stable equilibrium. (6) Discuss the relationship
of the result in (a) to Problem 2.37, page 53.
2.106. (a) Prove that a sufficient condition for the point (a,b) to be a minimum point of the function
V(x, y) is that at (a, b)
(i)
° V = d Z=0, (ii) . ^ (SrYSUS^ > and^>0
to = W = "' w " VteVV* 8 / \ s »*
(6) Use (a) to investigate the points of stability of a particle moving in a force field having potential
y = jj.3 + y z  33. _ \2y. Ans. (b) The point (1, 2) is a point of stability
2.107. Suppose that a particle of unit mass moves in the force field of Problem 2.106. Find its speed
at any time.
2.108. A particle moves once around the circle r = a(cos B i + sin e j) in a force field
F = (xiyj)/(x 2 + y 2 )
(a) Find the work done. (6) Is the force field conservative? (c) Do your answers to (a) and (6)
contradict Theorem 2.4, page 35? Explain.
2.109. It is sometimes stated that classical or Newtonian mechanics makes the assumption that space and
time are both absolute. Discuss what is meant by this statement.
ft 2
J Fdt
2.110. The quantity F av = I 1 _ . is called the average force acting on a particle from time t x to t 2 .
2 1
Does the result (3) of Problem 2.5, page 40, hold if F is replaced by F av ? Explain.
CHAP. 2] NEWTON'S LAWS OF MOTION. WORK, ENERGY AND MOMENTUM 61
2.111. A particle of mass 2 gm moves in the force field F = Sxyi + (4x 2 — 8z)j — Syk. dynes. If it has
a speed of 4 cm/sec at the point (—1, 2, —1), what is its speed at (1, —1, 1)? Ans. 6 cm/sec
2.112. (a) Find positions of stable equilibrium of a particle moving in a force field of potential
V = 18r 2 e2r.
(o) If the particle is released at r = £, find the speed when it reaches the equilibrium position,
(c) Find the period for small oscillations about the equilibrium position.
2.113. According to Einstein's special theory of relativity the mass m of a particle moving with speed
v relative to an observer is given by m = m /Vl — v 2 /c 2 where c is the speed of light [186,000
mi/sec] and m is the rest mass. What is the percent increase in rest mass of (a) an airplane moving
at 700 mi/hr, (6) a planet moving at 25,000 mi/hr, (c) an electron moving at half the speed of light?
What conclusions do you draw from these results?
2.114. Prove that in cylindrical coordinates,
dV ldV , dV
dp e " + P 30 ** dz
where e p , e^, e z are unit vectors in the direction of increasing p, <f> and z respectively.
VF = ^e p + i ^e* + ^e,
2.115. Prove that in spherical coordinates,
T7T/ dV . ! dV _l 1 3V
Vy = ^ + r Jt e ° + ^In7"s7 e *
where e r , e e , e^ are unit vectors in the direction of increasing r, e, <f> respectively.
UNIFORM FORCE FIELDS
A force field which has constant magni
tude and direction is called a uniform or con
stant force field. If the direction of this field
is taken as the negative z direction as indi
cated in Fig. 31 and the magnitude is the
constant F > 0, then the force field is given by
F = F k {1)
urn
F«k
Fig. 31
UNIFORMLY ACCELERATED MOTION
If a particle of constant mass m moves in a uniform force field, then its acceleration
is uniform or constant. The motion is then described as uniformly accelerated motion.
Using F = ma in (1), the acceleration of a particle of mass m moving in the uniform force
field (1) is given by „
a = ^k (*)
m v
WEIGHT AND ACCELERATION DUE TO GRAVITY
It is found experimentally that near the earth's
surface objects fall with a vertical acceleration
which is constant provided that air resistance is
negligible. This acceleration is denoted by g and
is called the acceleration due to gravity or the
gravitational acceleration. The approximate mag
nitude of g is 980 cm/sec 2 , 9.80 m/sec 2 or 32 ft/sec 2
according as the cgs, mks or fps system of units
is used. This value varies at different parts of the
earth's surface, increasing slightly as one goes
from the equator to the poles.
Assuming the surface of the earth is repre
sented by the xy plane of Fig. 32, the force acting
on a particle of mass m is given by
W = — mgk
(3)
This force, which is called the weight of the par
ticle, has magnitude W — mg.
Fig. 32
62
CHAP. 3] MOTION IN A UNIFORM FIELD. FALLING BODIES AND PROJECTILES 63
GRAVITATIONAL SYSTEM OF UNITS
Because W = mg, it follows that m = Wig. This fact has led many scientists and engi
neers, who deal to a large extent with mechanics on the earth's surface, to rewrite the
equations of motion with the fundamental mass quantity m replaced by the weight quantity
W. Thus, for example, Newton's second law is rewritten as
W
F = fa W
In this equation W and g can both vary while m = Wig is constant. One system of
units used in (4) is the gravitational or English engineering system where the unit of F or
W is the pound weight (lb wt) while length is in feet and time is in seconds. In this case
the unit of m is the slug and the system is often called the footslug second (fss) system.
Other systems are also possible. For example, we can take F or W in kilograms weight
(kg wt) with length in meters and time in seconds.
ASSUMPTION OF A FLAT EARTH
Equation (3) indicates that the force acting on mass m has constant magnitude mg and
is at each point directed perpendicular to the earth's surface represented by the xy plane.
In reality this assumption, called the assumption of the flat earth, is not correct first because
the earth is not flat and second because the force acting on mass m actually varies with the
distance from the center of the earth, as shown in Chapter 5.
In practice the assumption of a flat earth is quite accurate for describing motions of
objects at or near the earth's surface and will be used throughout this chapter. However,
for describing the motion of objects far from the earth's surface the methods of Chapter 5
must be employed.
FREELY FALLING BODIES
If an object moves so that the only force acting upon it is its weight, or force due to
gravity, then the object is often called a freely falling body. If r is the position vector and
m is the mass of the body, then using Newton's second law the differential equation of
motion is seen from equation (3) to be
d 2 r , d*r
m dT> = ~ mgk or d¥ = ~ 9k (*)
Since this equation does not involve the mass m, the motion of a freely falling body is
independent of its mass.
PROJECTILES
An object fired from a gun or dropped from a moving airplane is often called a projectile.
If air resistance is negligible, a projectile can be considered as a freely falling body so that
its motion can be found from equation (5) together with appropriate initial conditions. If air
resistance is negligible the path of a projectile is an arc of a parabola (or a straight line
which can be considered a degenerate parabola). See Problem 3.6.
64 MOTION IN A UNIFORM FIELD. FALLING BODIES AND PROJECTILES [CHAP. 3
POTENTIAL AND POTENTIAL ENERGY
IN A UNIFORM FORCE FIELD
The potential of the uniform force field, or potential energy of a particle in this force
field, is given by , v
V = Fo(zzo) (6)
where z is an arbitrary constant such that when z = z , V = 0. We call z = z the reference
level.
In particular for a constant gravitational field, F = mg and the potential energy of the
particle is
V = mg(zz ) v)
This leads to
Theorem 3.1. The potential energy of a particle in a constant gravitational field is
found by multiplying the magnitude of its weight by the height above some prescribed
reference level. Note that the potential energy is the work done by the weight in moving
through the distance z — zo.
MOTION IN A RESISTING MEDIUM
In practice an object is acted upon not only by its weight but by other forces as well.
An important class of forces are those which tend to oppose the motion of an object. Such
forces, which generally arise because of motion in some medium such as air or water, are
often called resisting, damping or dissipative forces and the corresponding medium is said
to be a resisting, damping or dissipative medium.
It is found experimentally that for low speeds the resisting force is in magnitude propor
tional to the speed. In other cases it may be proportional to the square [or some other power]
of the speed. If the resisting force is R, then the motion of a particle of mass m in an
otherwise uniform (gravitational) force field is given by
dt 2
If R = this reduces to (5).
m^ = mgk — R (#)
ISOLATING THE SYSTEM
In dealing with the dynamics or statics of a particle [or a system of particles, as we shall
see later] it is extremely important to take into account all those forces which act on the
particle [or on the system of particles]. This process is often called isolating the system.
CONSTRAINED MOTION
In some cases a particle P must move along some specified curve or surface as, for
example, the inclined plane of Fig. 33 or the inner surface of a hemispherical bowl of
Fig. 34 below. Such a curve or surface on which the particle must move is called a
constraint and the resulting motion is called constrained motion.
Just as the particle exerts a force on the constraint, there will by Newton's third law
be a reaction force of the constraint on the particle. This reaction force is often described
by giving its components N and f, normal to and parallel to the direction of motion
respectively. In most cases which arise in practice, f is the force due to friction and is
taken in a direction opposing the motion.
CHAP. 3] MOTION IN A UNIFORM FIELD. FALLING BODIES AND PROJECTILES
65
n\
>&^
A
s^ W
s^\ a
Fig. 33
Fig. 34
Problems involving constrained motion can be solved by using Newton's second law
to arrive at differential equations for the motion and then solving these equations subject to
initial conditions.
FRICTION
In the constrained motion of particles, one of the
most important forces resisting motion is that due to
friction. Referring to Fig. 35, let N be the magnitude
of the normal component of the reaction of the con
straint on the particle m. Then it is found experi
mentally that the magnitude of the force f due to
friction is given by
/ = /JV (P)
Fig. 35
where /* is called the coefficient of friction. The direction of f is always opposite to the
direction of motion. The coefficient of friction, which depends on the material of both
the particle and constraint, is taken as a constant in practice.
STATICS IN A UNIFORM GRAVITATIONAL FIELD
As indicated in Chapter 2, a particle is in equilibrium under the influence of a system of
forces if and only if the net force acting on it is F = 0.
Solved Problems
Fi
UNIFORM FORCE FIELDS AND UNIFORMLY ACCELERATED MOTION
3.1. A particle of mass m moves along a
straight line under the influence of a con
stant force of magnitude F. If its initial
speed is v , find (a) the speed, (b) the
velocity and (c) the distance traveled
after time t. Fig. 36
66 MOTION IN A UNIFORM FIELD. FALLING BODIES AND PROJECTILES [CHAP. 3
(a) Assume that the straight line along which the particle P moves is the x axis, as shown in
Fig. 36 above. Suppose that at time t the particle is at a distance x from origin 0. If i is a
unit vector in the direction OP and v is the speed at time t, then the velocity is vi. By Newton's
second law we have
4r(mvi) = Fi or m^j = F (1)
dt dt
Thus dv = —dt or J dv = J — dt
i.e. v = —t+C! {2)
m
where c^ is a constant of integration. To find c x we note the initial condition that v = v at
t = so that from (2), c x = v and
v = — t + v or v = v + — t (S)
(6) From (5) the velocity at time t is
If F
vi = v i H ti or v = v H 1
where v = vi, v = v i and F = Fi.
(c) Since v = dse/d* we have from (8),
* = * + £« » r *=(•, + £♦)*
Then on integrating, assuming c 2 to be the constant of integration, we have
* = * + (k)." + *
Since a; = at t = 0, we find c 2 = 0. Thus
• = * + (£)* w
3.2. Referring to Pr oblem 3.1, sho w that the speed of the particle at any position x is
given by v = V v l + (2F/m)x.
Method 1.
From (S) of Problem 3.1, we have t = m(v  v )/F. Substituting into (4) and simplifying,
we find x = (m/2F)(v 2  v„). Solving for v we obtain the required result.
Method 2.
From {1) of Problem 3.1, we have
dv _ F . dv dx _ F
dt ~ m' ' dx dt m
or since v = dx/dt,
„*1 = £, i. e . vdv = ?dx
dx m' m
v 2 _ F ,
Integrating, y  J£* + c 3
Since v = v when g = 0, we find c 3 = «g/2 and hence v = V«{j + (2F/m)x.
Method 3.
Change in kinetic energy from t = to any time t
= Work done in moving particle from * = to any position x
or $mv*  %mv% = F(x  0). Then v = V^ + (2F/m)x.
CHAP. 3] MOTION IN A UNIFORM FIELD. FALLING BODIES AND PROJECTILES
67
LINEAR MOTION OF FREELY
FALLING BODIES
3.3. An object of mass m is thrown vertically up
ward from the earth's surface with speed vo.
Find (a) the position at any time, (b) the time
taken to reach the highest point and (c) the
maximum height reached.
(a) Let the position vector of m at any time t be
r = xi + yj + zk. Assume that the object starts
at r = when t = 0. Since the force acting on
the object is — mgk, we have by Newton's law,
F =
Fig. 37
d 2 r
dt 2
dv .
mr = — mgk
dv
dt
= fl*
(J)
where v is the velocity at time t. Integrating (1) once yields
v = gtk + c t (2)
Since the velocity at t = [i.e. the initial velocity] is v<jk> we have from (2), e t = Vok so that
v = — gtk + V(jk = (v — gt)k
(S)
(*)
(5)
(6)
(7)
or ^ = (v gt)k
Integrating (4) yields r = (v t — %gt 2 )k + c 2
Then since r = when t = 0, c 2 = 0. Thus the position vector is
r = (v Q t$gt2)k
or, equivalently, x = 0, y = 0, z = v t — ^gt 2
(6) The highest point is reached when v = (v — gt)k = 0, i.e. at time t = vjg.
(c) At time t = VfJg the maximum height reached is, from (7), z = i%/2g.
Another method.
If we assume, as is physically evident, that the object must always be on the z axis, we may
avoid vectors by writing Newton's law equivalently as [see equation (1) above and place r = zk]
cPz/dt 2 = g
from which, using z = 0, dz/dt = v at t = 0, we find
z = v t  %gt 2
as above. The answers to (6) and (c) are then obtained as before.
3.4. Find the speed of the particle of Problem 3.3 in terms of its distance from origin O.
Method 1. From Problem 3.3, equations (3) and (7), we have
v = v  gt, z = v t — \gt 2
Solving for t in the first equation and substituting into the second equation, we find
2 _ «2
/v v\ /v v\ 2 v^v
= Vo \T)t g (T) = *
„2 — „2 _
2gz
Method 2. From equation (1) of Problem 3.3 we have, since v = vk and v = dz/dt,
dv dv dz __. .. dv
dt
dv dz _
dz dt ~ 9
V Tz = ~ g
= — g, i.e. — — = — g or
Then on integrating, vV2 = — gz + c 3 . Since v = v at z = 0, c 3 = v^/2 and thus v 2 = v 2 — 2gz.
Method 3. See Problem 3.9 for a method using the principle of conservation of energy.
68
MOTION IN A UNIFORM FIELD. FALLING BODIES AND PROJECTILES [CHAP. 3
MOTION OF PROJECTILES
3.5. A projectile is launched with initial speed
v at an angle a with the horizontal. Find
(a) the position vector at any time, (b) the
time to reach the highest point, (c) the
maximum height reached, (d) the time of
flight back to earth and (e) the range.
(a) Let r be the position vector of the projec
tile and v the velocity at any time t. Then
by Newton's law
d 2 r ,
i.e.,
dt 2
(1)
= flrk
Fig. 38
dv ,
(2)
Integration yields
v = gtk + c x W
Assume the initial velocity of the projectile is in the yz plane so that the initial velocity is
v = v cos a j + v sin a k (4)
Since v = v at t = 0, we find from (S),
v = v cos a j + (v sin a — gt)k (5)
Replacing v by dr/dt in (5) and integrating, we obtain
r = (v cos a)tj + {(v Q sin a)t %gt 2 }k (6)
or, equivalents, x = 0, # = (v cos a)*, 2 = (v sin a)*  \g& (7)
It follows that the projectile remains in the yz plane.
(b) At the highest point of the path the component of velocity v in the k direction is zero. Thus
Vq sin a
is the required time.
v Q sin a
gt = and t =
g
(c) Using the value of t obtained in (&), we find from (7) that
'v sin a
Maximum height reached = (v sin a)
9
y
t> sina\ 2 aj^sin 2 a
2g
(8)
(9)
(d) The time of flight back to earth is the time when z = 0, i.e. when
{v sin a)t  y& = t[(v Q sin a)  ±gt] =
or since t ¥* 0,
t =
2v sin a
<i0)
Note that this is twice the time in (6).
(e) The range is the value of y at the time given by (10), i.e.,
/ 2v sin a \ _
Range = (v cos a) ( ) 
2v sin a cos a
vl sin 2a
3.6. Show that the path of the projectile in Problem 3.5 is a parabola.
From the second equation of (7) in Problem 3.5, we have t = y/(v cosa). Substituting this
into the third equation of (7) in Problem 3.5, we find
z = {v sin a)(y/v cos a)  y{yh cos a) 2 or z = y tan «  (f//2 V 2 )2/ 2 sec 2 a
which is a parabola in the yz plane.
CHAP. 3] MOTION IN A UNIFORM FIELD. FALLING BODIES AND PROJECTILES
69
3.7. Prove that the range of the projectile of Problem 3.5 is a maximum when the launch
ing angle a = 45°.
By Problem 3.5(e) the range is (v§ sin 2a)/ g. This is a maximum when sin 2a = 1, i.e. 2a = 90°
or a = 45°.
POTENTIAL AND POTENTIAL ENERGY
IN A UNIFORM FORCE FIELD
3.8. (a) Prove that a uniform force field is conservative, (b) find the potential correspond
ing to this field and (c) deduce the potential energy of a particle of mass m in a
uniform gravitational force field.
(a) If the force field is as indicated in Fig. 31, then F = — F k. We have
V XF =
Thus the force field is conservative.
i J k
d/dx d/dy d/dz
F
=
dV. dV . dV. „, dV . dV dV „ . . . ,
— — i — — j  — k. Then — =0, — = 0, — = F from which
dx dy dz dx dy dz
(b) F = F k = VV
V = F Q z + c. If V = at z = z , then c = —F z and so V = F (z — z )
(c) For a uniform gravitational force field, F = mgls. [see Fig. 32, page 62] and corresponds
to F = mg. Then by part (b) the potential or potential energy is V = mg(z — z Q ).
3.9. Work Problem 3.4 using the principle of conservation of energy.
According to the principle of conservation of energy, we have
P.E. at z = + K.E. at z = = P.E. at z + K.E. at z
^mvl = mgz + mi) 2
+
Then v 2 = v*  2gz.
MOTION IN A RESISTING MEDIUM
3.10. At time t = a. parachutist [Fig. 39] having
weight of magnitude mg is located at z = and
is traveling vertically downward with speed vo.
If the force or air resistance acting on the
parachute is proportional to the instantaneous
speed, find the (a) speed, (b) distance traveled
and (c) acceleration at any time t > 0.
(a) Assume the parachutist (considered as a particle
of mass m) is located at distance z from origin O.
If k is a unit vector in the vertically downward
direction, then the weight is mgls. while the force
of air resistance is — (3vk so that the net force
is (mg — fiv)k.
Thus by Newton's law,
(mg
rark =
at
/3v)k
(1)
Fig. 39
70 MOTION IN A UNIFORM FIELD. FALLING BODIES AND PROJECTILES [CHAP. 3
dv „ mdv
i.e. wijt = wgr — Bv or r = dt
dt mg — Bv
Integrating,  7f ln ( mg ~ Bv ^ ~ l + c i (*)
AM
Since v = v at t = 0, c x = — — In (ra^r — Bv ). Then from (2),
t = jln(mgpv ) jln (mg/3v) = jm ( ^_^ j
Thus X7* = e " t/m or * = T + ( v °~ 2 f) e ~ &t/m {s)
(b) From (5), dz/dt = mg/B + (v — mg/B)eW m . Then by integration,
mpt m/ mg\ _ Mlm .
z = f jf^v ^je 3t/m + C2
Since z = at £ = 0, c 2 = (m/B)(v — mg/B) and thus
« = ^+7 (•*?) a •"•")
(c) From (5), the acceleration is given by
04)
• = % = £(»»f> e " m = ('S)'" <*>
3.11. Show that the parachutist of Problem 3.10 approaches a limiting speed given by mglp.
Method 1.
From equation (5) of Problem 3.10, v = mg/B + (v — mg/B)eW m . Then as t increases,
v approaches mg/B so that after a short time the parachutist is traveling with speed which is
practically constant.
Method 2.
If the parachutist is to approach a limiting speed, the limiting acceleration must be zero.
Thus from equation (1) of Problem 3.10 we have mg — Bv hm = or v lim = mg//3.
3.12. A particle of mass m is traveling along the x axis such that at t — it is located at x =
and has speed vo. The particle is acted upon by a force which opposes the motion
and has magnitude proportional to the square of the instantaneous speed. Find the
(a) speed, (b) position and (c) acceleration of the particle at any time t > 0.
(a) Suppose particle P is at a distance x from O at
* = and has speed v [see Fig. 310]. Then the F = Bv 2 i
force F = — Bv 2 i where B > is a constant of
proportionality. By Newton's law,
mrri = —Bv 2 i or 5 = — —dt (J)
dt ^ v 2 m
Integrating, — 1/v = —Bt/m + c v Since v = v
when t = 0, we have c x = 1/iv Thus Fig. 310
1 /3t 1 wv
= — or v = — — t— r —
v m v Bv t + m
which is the speed.
(6) From W, f = .r? . Then f i. = f „£** = =» f ■
dt
Bv t + m ' J J Bv t + m Bv J t + m/Bv
m ^ ( * , »\ ,
— or
CHAP. 3] MOTION IN A UNIFORM FIELD. FALLING BODIES AND PROJECTILES 71
Since x = at t = 0, c 2 = ^ln(^V Thus
• = M (+ ^)f ln te) = f'»( i+ ^)
(c) From (a),
_ £k _ d ( mv ° \ = PmVQ u\
a ~ dt dt\pv t + mj (f3v t + m)* w
Note that although the speed of the particle continually decreases, it never conies to rest.
3.13. Determine the (a) speed and (b) acceleration of the particle of Problem 3.12 as a
function of the distance x from O.
Method 1. From parts (a) and (6) of Problem 3.12,
m /fiv t + m\ mv /3v t + m v
In , and " — ~ —
P \ m ) ' fiv Q t + m m v
Then s=jln(^J or v = i> «** /m
and the acceleration is given in magnitude by
dv Pn _ Br/m dx P v l _ 2Bx/m
dt m dt m
which can also be obtained from equation (4.) of Problem 3.12.
Method 2. From equation (1) of Problem 3.12 we have
dv dv dx dv n „
m77 = mj 77 = mvj = — Bv 2
dt dx dt dx
or since v ¥= 0, m =— = — /8v and — = — —x. Integrating, lnv = —fix/m + c 3 . Since v = v
ax v m
when x = 0, c 3 = In v . Thus In (v/v ) = —Bx/m or v = VQe  ^/™.
3.14. Suppose that in Problem 3.5 we assume that the projectile has acting upon it a force
due to air resistance equal to — pv where /? is a positive constant and v is the instan
taneous velocity. Find (a) the velocity and (b) the position vector at any time.
(a) The equation of motion in this case is
cPt dv
mjg = — mgk — /?v or m ~Jt^~ P y = — w #k C)
( 0/m dt
Dividing by m and multiplying by the integrating factor e J = eP t/m , the equation can
be written as
±{ e fit/m y } = gefit/m^
Integration yields e Pt/m v = — ■£ e& t/m k + c x (2)
The initial velocity or velocity at t = is
Vo = v Q cos a j + v Q sin a k (8)
Using this in (2) we find
c i = v o c °s o] + v<> sin a k I — ^k
72
MOTION IN A UNIFORM FIELD. FALLING BODIES AND PROJECTILES [CHAP. 3
Thus (2) becomes on dividing by e^ t/m ,
v = (v cos a j + v sin a k)e~^ m — ^(1 — e~W™)k
(b) Replacing v by dr/dt in (4) and integrating, we find
m 9 <* jl m — <
Since r = at * = 0,
Using (6) in (5), we find
mv Q
r = ~"g( v 0COSa j + v sinak)e^"» f(t + ^ e flt/m)k + Ca
fi v /?
J8 2 '
r = ^(cosa j + sin a k)(l  e"t/m)  2M / t + ™ /3t/m _ 2?M k
(4)
(5)
(*)
(7)
3.15. Prove that the projectile of Problem 3.14 attains a limiting velocity and find its value.
Method 1.
Refer to equation (4) of Problem 3.14. As t increases, e~» t/m approaches zero. Thus the
velocity approaches a limiting value equal to v llm = —(mg/(3)k.
Method 2.
If the projectile is to approach a limiting velocity its limiting acceleration must be zero. Thus
from equation (1) of Problem 3.14, mgk  0v llm = or v lim = (mg/p)k.
CONSTRAINED MOTION
3.16. A particle P of mass m slides without rolling
down a f rictionless inclined plane AB of angle a
[Fig. 311]. If it starts from rest at the top A
of the incline, find (a) the acceleration, (b) the
velocity and (c) the distance traveled after
time t.
(a) Since there is no friction the only forces acting
on P are the weight W = — mgk and the re
action force of the incline which is given by the
normal force N.
Let e x and e 2 be unit vectors parallel and
perpendicular to the incline respectively. If we
denote by s the magnitude of the displacement
from the top A of the inclined plane, we have
by Newton's second law
— *
N /
N. pjf /■ inff si" a «i
<^.
W = mgk
— mg cos a e 2
Fig. 311
d 2
mjzisei) = W + N = raflrsinae!
<J>
since the resultant equal to W + N is mg sin a e^ as indicated in Fig. 311. From (I) we have
d?s/d&  g sin a (2)
Thus the acceleration down the incline at any time t is a constant equal to g sin a.
(6) Since v = ds/dt is the speed, (2) can be written
dv/dt = flrsina or v — (g sin a)t + c t
on integrating. Using the initial condition v = at t = 0, we have c x = so that the
speed at any time t is
v = (g sin a)t (3)
The velocity is ve t = (g sin ajte^ which has magnitude (g sin a)t in the direction e t down
the incline.
CHAP. 3] MOTION IN A UNIFORM FIELD. FALLING BODIES AND PROJECTILES
73
(c) Since v = ds/dt, (5) can be written
ds/dt = (g sin a)t or s = %(9 sin a)t 2 + c 2
on integrating. Using the initial condition s = at t = 0, we find c 2 = so that the
required distance traveled is ,, . w , /»\
8 = (p sin a)t 2 W
3.17. If the length AB of the incline in Problem 3.16 is I, find (a) the time r taken for the
particle to reach the bottom B of the incline and (&) the speed at B.
(a) Since 8 = 1 at B, the time t to reach the bottom is from equation (4) of Problem 3.16 given by
I = ^(g sin a )r 2 or r  }/2l/(g sin a).
(6) The speed at £ is given from (5) of Problem 3.16 by v = (g sin «)r = V2tf* sin a.
MOTION INVOLVING FRICTION
3.18. Work Problem 3.16 if the inclined plane has
a constant coefficient of friction ju..
(a) In this case there is, in addition to the forces W
and N acting on P, a frictional force f [see Fig.
312] directed up the incline [in a direction oppo
site to the motion] and with magnitude
liN = v.mg cos a CO
i.e.
pmg cos a
i = —ping cos a ^
(2)
N>
f f s mg sin a ti
vjW
W = mgk
' s \ / e 2
*) ^>V
— mg cos a e 2 ' a /\^
Fig. 312
Then equation (1) of Problem 3.16 is replaced by
cP(*e t )
= W + N + f = mflr sin a x e x — y.mg cos a ej
m
dt 2
(S)
or <Ps/dt* = flr(sin a  /* cos a) W
Thus the acceleration down the incline has the constant magnitude g(sin a  ? cos a) provided
sin a > /icosa or tana > /* [otherwise the frictional force is so great that the particle will
not move at all].
(6) Replacing d 2 s/dt 2 by dv/dt in U) and integrating as in part (6) of Problem 3.16, we find the
speed at any time t to be (K x
v = gr( S in a — /t COS a)t l°J
(c) Replacing v by <*«/<** in (5) and integrating as in part (c) of Problem 3.16, we find
8 = ^g(sina — (i cos a)t 2
(6)
3.19. An object slides on a surface of ice along the horizontal straight line OA [Fig. 313].
At a certain point in its path the speed is v and the object then comes to rest after
traveling a distance xo. Prove that the coefficient of friction is v 2 /2gx .
Let * be the instantaneous distance of the
object of mass m from O and suppose that at
time t = 0, x = and dx/dt = v .
Three forces act on the object, namely (1) the
weight W = mg, (2) the normal force N of the
ice surface on the object, and (3) the frictional
force f.
O
mg
Fig. 313
By Newton's second law we have, if v is the instantaneous speed,
w ^i = W + N + f
dt
(1)
74
MOTION IN A UNIFORM FIELD. FALLING BODIES AND PROJECTILES [CHAP. 3
But since N = W and the magnitude of f is f = pN = ymg so that £ = —fimgi, (1) becomes
dv
dt
dv . . dv
m^i = fimgi or ^ = fig
Method 1. Write (2) as
dv dx dv
dx"di = m or v Tx = **
Then v dv = —figdx
Integrating, using the fact that v = v at x = 0, we find
v*/2 = mx + v\l1
Then since v = when x = x , (4) becomes
figx Q + v*/2 = or n = v*/2gx Q
(4)
(5)
Method 2. From (2) we have, on integrating and using the fact that v = v at t = 0,
v = v — figt or dx/dt = v — figt (6)
Integrating again, using the fact that x = at t = 0, we find
x = v t frgfi (7)
From (7) we see that the object comes to rest (i.e., v = 0) when
v — ngt = or t — v //xg
Substituting this into (7) and noting that x = x , we obtain the required result.
STATICS IN A UNIFORM GRAVITATIONAL FIELD
3.20. A particle of mass m is suspended in equilibrium by two inelastic strings of lengths
a and b from pegs A and B which are distant c apart. Find the tension in each string.
L
— mgk
Fig. 314
Fig. 315
Let W denote the weight of the particle and T t and T 2 the respective tensions in the strings
of lengths a and b as indicated in Fig. 314. These forces are also indicated in Fig. 315 and are
assumed to lie in the plane of unit vectors j and k. By resolving T 1 and T 2 into horizontal and
vertical components it is clear that
= T x sin a k — T x cos a j,
T 2 = T 2 sin p k + T 2 cos ft j
where T 1 and T 2 are the magnitudes of T x .and T 2 respectively and where a and /J are the respective
angles at A and B. Also we have
W = — mgk.
Since the particle is in equilibrium if and only if the net force acting on it is zero, we have
F = T x + T 2 + W
= 2 1 ! sin a k — T x cos a j + T 2 sin /? k + T 2 cos /? j — mgk
= (T 2 cos p — T t cos a)j + (2\ sin a + T 2 sin p  mg)k
=
CHAP. 3] MOTION IN A UNIFORM FIELD. FALLING BODIES AND PROJECTILES
75
From this we must have
T 2 cos /3  T t cos a  0, T x sin a + T 2 sin ft  mg =
Solving simultaneously, we find
„ _ mg cos p j, _ my cos a
1 ~ sin (<* + £)' 2 sin (<* + /?)
The angles a and /? can be determined from the law of cosines as
a = COS 1
a 2 + c 2  b 2
2ac
b 2 + c 2  a 2
26c
From these the tensions can be expressed in terms of a, b, c.
MISCELLANEOUS PROBLEMS
3.21. An inclined plane [Fig. 316] makes an
angle a with the horizontal. A projectile is
launched from the bottom A of the incline
with speed vo in a direction making an
angle jB with the horizontal.
(a) Prove that the range R up the incline is
given by
2v 2 sin (/? — a) cos /?
R =
g COS z a
Fig. 316
(b) Prove that the maximum range up the incline is given by
■Umax —
and is achieved when /? = tt/4 + a/2.
g(l + sin a)
(a) As in Problem 3.5, equation (6), the position vector of the projectile at any time t is
r = (v cos/?)«j + {(v Q sin p)t  lgt 2 }k (1)
or y = (v cos /?)*, z = (v sin /?)£  \gt 2 (2)
The equation of the incline [which is a line in the yz plane] is
z = y tan a (3)
we see that the projectile's path and the incline intersect for those
(v sirt /3)t — \gt 2 — [(v cos (3)t] tana
2v (sin /? cos a — cos /3 sin a) 2v sin (/J — a)
Using equations (2) in
values of t where
t =
and
g cos a
g cos a
The value t = gives the intersection point A. The second value of t yields point B
which is the required point. Using this second value of t in the first equation of (2), we find
that the required range R up the incline is
R = y sec a = {v cos /3)
2v sin (/? — o]
flf COS a
2v sin (/3 — a) cos /?
cos 2 a
(b) Method 1. The range R can be written by using the trigonometric identity
sin A cos B = £{sin (A + B) + sin (A  B)}
R =
vl
g cos 2 a
{sin (2/3 — a) — sin a}
76
MOTION IN A UNIFORM FIELD. FALLING BODIES AND PROJECTILES [CHAP. 3
This is a maximum when sin (2/3  o) = 1, i.e. 20  a = T /2 or fi = a/2 + tt/4, and the
value of this maximum is
R
v,
g cos z a
(1 — sin a) =
2
#(1 — sin 2 a)
(1 — sin a) —
g(l + sin a)
Method 2.
The required result can also be obtained by the methods of differential calculus for
finding maxima and minima.
3.22. Two particles of masses mi and m 2 respectively are
connected by an inextensible string of negligible mass
which passes over a fixed frictionless pulley of negli
gible mass as shown in Fig. 317. Describe the motion
by finding (a) the acceleration of the particles and
(b) the tension in the string.
Let us first isolate mass m,. There are two forces acting
on it: (1) its weight m y g = m x gk, and (2) the force due to
the string which is the tension T = — Tk. If we call a = dk
the acceleration, then by Newton's law
m x ak = m t gk — Tk (1)
Next we isolate mass m 2 . There are two forces acting
on it: (1) its weight m 2 g = m^k, and (2) the tension
T = — Tk [the tension is the same throughout the string
since the mass of the string is assumed negligible and in
extensible]. Since the string is inextensible, the acceleration
of ra 2 is —a = — ak. Then by Newton's law
— m 2 ak = m 2 gk — Tk (2)
'mWWXW
Wig
m 2
m 2 g
Fig. 317
From (1) and (2) we have
m^a
Solving simultaneously, we find
m x g — T,
m 2 a
m 2 g
m i ~ m 2
mi + m 2
g,
T =
2m 1 m 2
m 1 + m 2 i
Thus the particles move with constant acceleration, one particle rising and the other falling.
In this pulley system, sometimes called Atwood's machine, the pulley can rotate. However,
since it is frictionless and has no mass [or negligible mass] the effect is the same as if the string
passed over a smooth or frictionless peg instead of a pulley. In case the mass of the pulley
is not negligible, rotational effects must be taken into account and are considered in Chapter 9.
3.23. A particle P of mass m rests at the top A of a
frictionless fixed sphere of radius b. The par
ticle is displaced slightly so that it slides (with
out rolling) down the sphere, (a) At what posi
tion will it leave the sphere and (b) what will
its speed be at this position?
The particle will slide down a circle of radius a
which we choose to be in the xy plane as indicated in
Fig. 318. The forces acting on the particle are:
(1) its weight W = —mgj, and (2) the reaction force
N of the sphere on the particle normal to the sphere.
Method 1.
(a) Let the position of the particle on the circle be
measured by angle and let v i and t be unit
vectors. Resolving W into components in direc
tions r x and $ lt we have as in Problem 1.43,
page 24,
Fig. 318
CHAP. 3] MOTION IN A UNIFORM FIELD. FALLING BODIES AND PROJECTILES 77
W = (W'r,)r, + (W •#!)#,
= (— wgrj • r^ + (— mg\ • f x)*! = mg sin o r x — mg cos ^
Also, N = Nt x
Using Newton's second law and the result of Problem 1.49, page 26, we have
F = ma = m[(r — re 2 )r x + (r'o + 2re)$ x ]
= W + N = (2V — mflr sin tf)^ — mg cose 9 X (1)
Thus ra(rr« 2 ) = Nmg sine, m(r'e + 2r'e) = mg cos e (2)
While the particle is on the circle (or sphere), we have r = b. Substituting this into (2),
—mbe 2 = N — mg sin 0, be = —g cos e (3)
Multiplying the second equation by e, we see that it can be written
d /e 2 \ d
b Tt\2j = *S (Bin#)
Integrating, be 2 /2 = — g sin + c 1# Now when * = tt/2, = so that c x = g and
6ff2 = 2gr(l  sin a) (4)
Substituting (4) into the first equation of (S), we find
N = mg(Z sin e  2) (5)
Now as long as N > the particle stays on the sphere; but when N = the particle will
be just about to leave the sphere. Thus the required angle is given by 3 sin e — 2 = 0, i.e.,
sin* = 2/3 or e = sin" 1 2/3 (6)
(6) Putting sin 6 = £ into (4), we find
e 2 = 2^/36 (7)
Then if v is the speed, we have v = be so that (7) yields v 2 = f&p or v = yj\bg.
Method 2. By the conservation of energy, using the x axis as reference level, we have
P.E. at A + K.E. at A = P.E. at P + K.E. at P
mgb +0 = mgb sin e + fymv 2
or v 2 = 2gb(l  sin e) («)
Using the result of Problem 1.35, page 20, together with Newton's second law, we have, since
the radius of curvature is 6,
» = «• = Cf.I') = w+n
= (N — mg sin e)r x — mg cos e 6 X
Using only the r t component, we have
v 2 /b = N — mg sine (5)
From (8) and (9) we find N = mg(3 sin e  2) which yields the required angle sin" 1 (§) as in
Method 1. The speed is then found from (8).
78 MOTION IN A UNIFORM FIELD. FALLING BODIES AND PROJECTILES [CHAP. 3
Supplementary Problems
UNIFORM FORCE FIELDS AND LINEAR MOTION OF
FREELY FALLING BODIES
3.24. An object of mass m is dropped from a height H above th e grou nd. Prove that if air resistance
is negligible, then it will reach the ground (a) in a time y/2H/g and (b) with speed y/2gH.
3.25. Work Problem 3.24 if the object is thrown vertically downward with an initial velocity of magni
tude v . Ans. (a) (y/v* + 2gH  v )/g, (b) y/v* + 2gH
3.26. Prove that the object of Problem 3.3, page 67, returns to the earth's surface (a) with the same
speed as the initial speed and (6) in a time which is twice that taken to reach the maximum height.
3.27. A ball which is thrown upward reaches its maximum height of 100 ft and then returns to the
starting point, (a) With what speed was it thrown? (6) How long does it take to return?
Ans. (a) 80 ft/sec, (6) 5 sec
3.28. A ball which is thrown vertically upward reaches a particular height H after a time t x on the
way up and a time t 2 on the way down. Prove that (a) the initial velocity with which the ball was
thrown has magnitude $g(r t + t 2 ) and (6) the height H = $griT 2 .
3.29. In Problem 3.28, what is the maximum height reached? Ans. ^g(r t + t 2 ) 2
3.30. Two objects are dropped from the top of a cliff of height H. The second is dropped when the first
has traveled a distance D. Prove that at the instant w hen t he first object has reached the bottom,
the second object is at a distance above it given by 2yDH — D.
3.31. An elevator starts from rest and attains a speed of 16 ft/sec in 2 sec. Find the weight of a
160 lb man in the elevator if the elevator is (a) moving up (6) moving down.
Ans. (a) 200 lb, (6) 120 lb
3.32. A particle of mass 3 kg moving in a straight line decelerates uniformly from a speed of 40 m/sec
to 20 m/sec in a distance of 300 m. (a) Find the magnitude of the deceleration. (6) How much
further does it travel before it comes to rest and how much longer will this take?
Ans. (a) 2 m/sec 2 , (6) 100 m; 10 sec
3.33. In Problem 3.32, what is the total work done in bringing the particle to rest from the speed
of 40 m/sec? Ans. 2400 newton meters (or joules)
MOTION OF PROJECTILES
3.34. A projectile is launched with a muzzle velocity of 1800 mi/hr at an angle of 60° with a
horizontal and lands on the same plane. Find (a) the maximum height reached, (6) the time
to reach the maximum height, (c) the total time of flight, (d) the range, (e) the speed after
1 minute of flight, (/) the speed at a height of 32,000 ft.
Ans. (a) 15.5 mi, (6) 71.4 sec, (c) 142.8 sec, (d) 35.7 mi, (e) 934 mi/hr, (/) 1558 mi/hr
3.35. (a) What is the maximum range possible for a projectile fired from a cannon having muzzle
velocity 1 mi/sec and (6) what is the height reached in this case?
Ans. (a) 165 mi, (6) 41.25 mi
3.36 A cannon has its maximum range given by R max . Prove that (a) the height reached in such case
is £# max and (&) the time of flight is ^R m& J2g.
3.37. It is desired to launch a projectile from the ground so as to hit a given point on the ground
which is at a distance less than the maximum range. Prove that there are two possible angles
for the launching, one which is less than 45° by a certain amount and the other greater than
45° by the same amount.
3.38. A projectile having horizontal range R reaches a maximum height H . Prove that it must have
been launched with (a) an initial speed equal to y/g(R 2 + 16H 2 )/8H and (6) at an angle with
the horizontal given by sin 1 (4H/y/R 2 + 16Jf? 2 ).
CHAP. 3] MOTION IN A UNIFORM FIELD. FALLING BODIES AND PROJECTILES 79
3.39. A projectile is launched at an angle a from a
cliff of height H above sea level. If it falls into
the sea at a distance D from the base of the
cliff, prove that its maximum height above sea
level is „ _,_ Ctan'q
H+ 4(H + Ptan«) Fl S 3 " 19
MOTION IN A RESISTING MEDIUM
3.40. An object of weight W is thrown vertically upward with speed v . Assuming that air resistance
is proportional to the instantaneous velocity and that the constant of proportionality is k, prove
that (a) the object will reach a maximum height of
Wkv W 2 ln ( 1 + KV o
K 2 g K 2 g \ W
and that (6) the time taken to reach this maximum height is
Kg \ W )
3.41. A man on a parachute falls from rest and acquires a limiting speed of 15 mi/hr. Assuming
that air resistance is proportional to the instantaneous speed, determine how long it takes to reach
the speed of 14 mi/hr. Arts. 1.86 sec
3.42. A mass m moves along a straight line under the influence of a constant force F. Assuming
that there is a resisting force numerically equal to kv 2 where v is the instantaneous speed and k
w . /Fkv\
is a constant, prove that the distance traveled in going from speed v x to v 2 is ^ In ( F _ ^
3.43. A particle of mass m moves in a straight line acted upon by a constant resisting force of magni
tude F. If it starts with a speed of v , (a) how long will it take before coming to rest and
(6) what distance will it travel in this time? Ana. (a) mv /F, (b) mv 2 /2F
3.44. Can Problem 3.43 be worked by energy considerations? Explain.
3.45. A locomotive of mass m travels with constant speed v along a horizontal track, (a) How long
will it take for the locomotive to come to rest after the ignition is turned off, if the resistance
to the motion is given by a + /3v 2 where v is the in stant aneous speed and a and /3 are constants?
(b) What is the distance traveled? Ana. (a) y/mlp tan" 1 (v Q VJ/Z), (6) (m/2/3) In (1 + prf/a)
3.46. A particle moves along the x axis acted upon only by a resisting force which is proportional
to the cube of the instantaneous speed. If the initial speed is v and after a time r the speed is
£v , prove that the speed will be ±v in time 5t.
3.47. Find the total distance traveled by the particle of Problem 3.46 in reaching the speeds (a) v >
(6) £v . Ana. (a) §v r, (6) iv
3.48. Prove that for the projectile of Problem 3.14, page 71,
m f /?v sina\
(a) the time to reach the highest point is j In ( 1 H — — ) ,
mv sina m 2 g , / J»oraa\
(6) the maximum height is tj p In I 1 H — ) .
CONSTRAINED MOTION AND FRICTION
3.49. A weight of 100 lb slides from rest down a 60° incline of length 200 ft starting from the top.
Neglecting friction, (a) how long will it take to reach the bottom of the incline and (6) what is
the speed with which it reaches the bottom? Ana. (a) 3.80 sec, (&) 105.3 ft/sec
80
MOTION IN A UNIFORM FIELD. FALLING BODIES AND PROJECTILES [CHAP. 3
3.50. Work Problem 3.49 if the coefficient of friction is 0.3. Ans. (a) 4.18 sec, (6) 95.7 ft/sec
3.51. (a) With what speed should an object be thrown up a smooth incline of angle a and length I,
starting from the bottom, so as to just reach the top and (6) what is the time taken?
Ans. (a) y/2gl sin a, (b) s/2l/(g sin a)
3.52. If it takes a time t for an object starting from speed v on an icy surface to come to rest, prove
that the coefficient of friction is v Q /gr.
3.53. What force is needed to move a 10 ton truck with uniform speed up an incline of 30° if the
coefficient of friction is 0.1 1 Ans. 5.87 tons
3.54. A mass m rests on a horizontal piece of wood. The wood is tilted upward until the mass m just
begins to slide. If the angle which the wood makes with the horizontal at that instant is a,
prove that the coefficient of friction is n = tan a.
3.55. A 400 kg mass on a 30° inclined plane is acted upon by a force of 4800 newtons at angle 30°
with the incline, as shown in Fig. 320. Find the acceleration of the mass if the incline (a) is
frictionless, (6) has coefficient of friction 0.2. Ans. (a) 5.5 m/sec 2 , (6) 5.0 m/sec 2
Fig. 320
Fig. 321
3.56. Work Problem 3.55 if the force of 4800 newtons acts as shown in Fig. 321.
Ans. (a) 5.5 m/sec 2 , (6) 2.6 m/sec 2
STATICS IN A UNIFORM GRAVITATIONAL FIELD
3.57. A 100 kg weight is suspended vertically from the center of a rope as shown in Fig. 322.
Determine the tension T in the rope. Ans. T = 100 kg wt = 980 nt
y///////////////^^^^^
Fig. 323
Fig. 324
3.58. In Fig. 323, AB and AC are ropes attached to the ceiling CD and wall BD at C and B respectively.
A weight W is suspended from A. If the ropes AB and AC make angles e x and e 2 with the
wall and ceiling respectively, find the tensions T t and T 2 in the ropes.
W cos 6 2 _ W sin e x
T 2 —
Ans. T, =
cos (0x — e 2 ) '
cos (e 1 — e 2 )
3.59. Find the magnitude of the force F needed to keep mass m in equilibrium on the inclined plane
of Fig. 324 if (a) the plane is smooth, (6) the plane has coefficient of friction p.
. . . _ mg sin a ... „ __ mg (sin a — fi cos a)
AnS ' {a) F = I^T' (6) F c^sl
CHAP. 3] MOTION IN A UNIFORM FIELD. FALLING BODIES AND PROJECTILES 81
3.60. How much force is needed to pull a train weighing 320 tons from rest to a speed of 15 mi/hr
in 20 seconds if the coefficient of friction is 0.02 and (a) the track is horizontal, (6) the track is
inclined at an angle of 10° with the horizontal and the train is going upward? [Use sin 10° = .1737,
cos 10° = .9848.] Ans. (a) 17.4 tons, (6) 129.6 tons
3.61. Work Problem 3.60(6) if the train is going down the incline. Ans. 3.6 tons
3.62. A train of mass m is coasting down an inclined plane of angle a and coefficient of friction fi
with constant speed v Q . Prove that the force needed to stop the train in a time t is given by
Wfl^sin a — (i cos a) + mt> /r.
MISCELLANEOUS PROBLEMS
3.63. A stone is dropped down a well and the sound of the splash is heard after time t. A ssuming the
speed of sound is c, prove that the depth of the water level in the well is (y/c 2 + Igcr — c) 2 /2g.
3.64. A projectile is launched downward from the top of an inclined plane of angle a in a direction
making an angle y with the incline. Assuming that the projectile hits the incline, prove that
2i>o sin y cos (y — a) ,,,,., . , .,
(a) the ranee is given by R = 5 and that (6) the maximum range down the
v ' 9 9 cos 2 a
K
g(l — sin a)
incline is i? max — ~" n —
3.65. A cannon is located on a hill which has the shape of an inclined plane of angle a with the horizontal.
A projectile is fired from this cannon in a direction up the hill and making an angle /3 with it. Prove
/ 2 sin 2a *
that in order for the projectile to hit the hill horizontally we must have = tan M 3 _ cog 2a
3.66. Suppose that two projectiles are launched at angles a and /3 with the horizontal from the
same place at the same time in the same vertical plane and with the same initial speed. Prove
that during the course of the motion, the line joining the projectiles makes a constant angle
with the vertical given by £(a + /3).
3.67. Is it possible to solve equation (1), page 33, by the method of separation of variables? Explain.
3.68. When launched at angle 9 1 with the horizontal a projectile falls a distance D x short of its target,
while at angle 2 it falls a distance D 2 beyond the target. Find the angle at which the projectile
should be launched so as to hit the target.
3.69. An object was thrown vertically downward. During the tenth second of travel it fell twice as far
as during the fifth second. With what speed was it thrown? Ans. 16 ft/sec
3.70. A gun of muzzle speed v is situated at height h above a horizontal plane. Prove that the angle
at which it must be fired so as to achieve the greatest range on the plane is given by
e = % cos 1 gh/(vl + gh).
3.71. In Fig. 325, AB is a smooth table and masses ra x
and w 2 are connected by a string over the smooth —
peg at B. Find (a) the acceleration of mass m 2
and (6) the tension in the string.
m 2 — m x
AnS  (o) ^T+nT 2 g ' m2>mi
jm^ Fig>3 . 25
Wi + m 2
3.72. Work Problem 3.71 if the table AB has coefficient of friction ft.
3.73. The maximum range of a projectile when fired down an inclined plane is twice the maximum
range when fired up the inclined plane. Find the angle which the incline makes with the horizontal.
Ans. sin 1 1/3
I I JJ
B
m 2
82
MOTION IN A UNIFORM FIELD. FALLING BODIES AND PROJECTILES [CHAP. 3
3.74. Masses m^ and ra 2 are located on smooth inclined planes
of angles a t and a 2 respectively and are connected by
an inextensible string of negligible mass which passes
over a smooth peg at A [Fig. 326]. Find the accelera
tions of the masses.
Ans. The accelerations are in magnitude equal to
Wi sin a x — m 2 sin a 2
m x + m 2
9
Fig. 326
3.75. Work Problem 3.74 if the coefficient of friction between the masses and the incline is ju.
Ans.
Wj sin <*! — m 2 sin a 2 — nm x cos a x — [im 2 cos a 2
mi + m 2
3.76. Prove that the least horizontal force F needed to pull
a cylinder of radius a and weight W over an obstacle
of height b [see Fig. 327] is given in magnitude by
Wy/b(2ab)/(a  6).
3.77. Explain mathematically why a projectile fired from
cannon A at the top of a cliff at height H above the
ground can reach a cannon B located on the ground,
while a projectile fired from cannon B with the same
muzzle velocity will not be able to reach cannon A.
3.78. In Fig. 328 the mass m hangs from an inextensible string OA.
It is pulled aside by a horizontal string AB so that OA makes
an angle a with the vertical. Find the tension in each string.
Ans. Tension in AB = mg tan a; in OA = mg sec a
3.79. A particle moving along the x axis is acted upon by a resisting
force which is such that the time t for it to travel a distance x
is given by t = Ax 2 + Bx + C where A, B and C are constants.
Prove that the magnitude of the resisting force is proportional
to the cube of the instantaneous speed.
Fig. 327
y////////////.
O
3.80.
3.81.
3.82.
3.83.
3.84.
A projectile is to be launched so as to go from A to B
[which are respectively at the bases of a double inclined
plane having angles a and /3 as shown in Fig. 329] and
just barely miss a pole of height H. If the distance
between A and B is D, find the angle with the horizontal
at which the projectile should be launched.
A particle of mass m moves on a frictionless inclined
plane of angle a and length I. If the particle starts
from rest at the top of the incline, what will be its
speed at the bottom assuming that air resistance is equal
to kv where v is the instantaneous speed and k is con
stant?
Fig. 329
Suppose that in Problem 3.23 the particle P is given an initial speed v Q at the top of the circle
(or sphere), (a) Prove that if v ^ yfgb, the angle e at which the particle leaves the circle is given
by sin 1 (§ + v^/Zgb). (b) Discuss what happens if v Q > ygb.
A cannon is situated at the top of a vertical cliff overlooking the sea at height H above sea level.
What should be the least muzzle velocity of the cannon in order that a projectile fired from it
will reach a ship at distance D from the foot of the cliff?
In Problem 3.83, (a) how long would it take the projectile to reach the ship and (6) what is the
velocity on reaching the ship?
CHAP. 3] MOTION IN A UNIFORM FIELD. FALLING BODIES AND PROJECTILES
83
3.85. A uniform chain of total length a has a portion
< b < a hanging over the edge of a smooth table
AB [see Fig. 330]. Prove that the time taken for
the chain to slide off the table if it starts from rest
is y/a/g\n(a + y/a 2 b 2 )/b.
Fig. 330
3.86. If the table in Problem 3.85 has coefficient of friction p, prove that the time taken is
In
V
a
(i + fig
a + V« 2  [6(1 + fi~ H 2
6(1 + n) — an
3.87. A weight W x hangs on one side of a smooth fixed pulley of neg
ligible mass [see Fig. 331]. A man of weight W 2 pulls himself
up so that his acceleration relative to the fixed pulley is a.
Prove that the weight W^ moves upward with acceleration given by
[9(W 2 W 1 )W 2 a]/W 1 .
3.88. Two monkeys of equal weight are hanging from opposite ends of a
rope which passes over a smooth fixed pulley of negligible mass.
The first monkey starts to climb the rope at a speed of 1 ft/sec
while the other remains at rest relative to the rope. Describe the
motion of the second monkey.
Ans. The second monkey moves up at the rate of 1 ft/sec.
W////////A
Fig. 331
3.89. Prove that the particle of Problem 3.23 will land at a distance from the base of the sphere
given by (4\^90 + 19^)6/81.
3.90. Prove that if friction is negligible the time taken for a particle to slide down any chord of a
vertical circle starting from rest at the top of the circle is the same regardless of the chord.
3.91. Given line AB of Fig. 332 and point P where AB and
P are in the same vertical plane. Find a point Q on
AB such that a particle starting from point P will
reach Q in the shortest possible time.
[Hint. Use Problem 3.90.]
3.92. Show how to work Problem 3.91 if line AB is re
placed by a plane curve. Can it be done for a space
curve? Explain.
3.93. Find the work done in moving the mass from the top of the incline of Problem 3.18 to the bottom.
Ans. mgl(sm a — n cos a)
3.94. The force on a particle having electrical charge q and which is moving in a magnetic field of intensity
or strength B is given by F = g(v X B) where v is the instantaneous velocity. Prove that if the
particle is given an initial speed v in a plane perpendicular to a magnetic field B of constant
strength, then it (a) will travel with constant speed v Q and (6) will travel in a circular path
of radius mv /qB. Assume that gravitational forces are negligible.
3.95. Prove that the period, i.e. the time for one complete vibration, of the particle of Problem 3.94
is independent of the speed of the particle and find its value. Ans. 2irm/qB
3.96. Work Problem 3.94 if B is constant and the particle is given an initial speed v in a plane which
is not necessarily perpendicular to the magnetic field. Can we define a period in this case? Explain.
84
MOTION IN A UNIFORM FIELD. FALLING BODIES AND PROJECTILES [CHAP. 3
3.97. If a particle of electrical charge q and mass m moves with velocity v in an electromagnetic field
having electric intensity E and magnetic intensity B the force acting on it, called the Lorentz force,
is given by
F = g(E + vXB)
Suppose that B and E are constant and in the directions of the negative y and positive z axes
respectively. Prove that if the particle starts from rest at the origin, then it will describe a
cycloid in the yz plane whose equation is
y = b(e — sin e), z = 6(1 — cos e)
where = qBt/m, b = mE/qB 2 and t is the time.
3.98. (a) An astronaut of 80 kg wt on the earth takes oft" vertically in a space ship which achieves
a speed of 2000 km/hr in 2 minutes. Assuming the acceleration to be constant, what is his apparent
weight during this time? (b) Work part (a) if the astronaut has 180 lb wt on the earth and the
space ship achieves a speed of 1280 mi/hr in 2 minutes. Ans. (a) 117 kg wt, (6) 268 lb wt
3.99.
In Problem 3.82, how far from the base of the sphere will the particle land?
3.100. In Fig. 333 weight W t is on top of weight W 2 which is in turn on a horizontal plane. The
coefficient of friction between W x and W 2 is ^ while that between W 2 and the plane is fi 2 . Suppose
that a force F inclined at angle a to the horizontal is applied to weight W^. Prove that if
cot a ^ fii > n 2 , then a necessary and sufficient condition that W 2 move relative to the plane
while Wi not move relative to W 2 is that
COS a — n 2 sin a
< F g
COS a — jii Sin a
Fig. 333
3.101. Discuss the results in Problem 3.100 if any of the conditions are not satisfied.
3.102. Give a generalization of Problem 3.100.
3.103. Describe the motion of the particle of Problem 3.97 if E and B are constants, and have the same
direction.
3.104. A bead of mass m is located on a parabolic wire with its axis
vertical and vertex directed downward as in Fig. 334 and
whose equation is cz = x 2 . If the coefficient of friction is
fi, find the highest distance above the x axis at which the
particle will be in equilibrium. Ans. \p?c
3.105. Work Problem 3.104 if the parabola is replaced by a vertical
circle of radius 6 which is tangent to the x axis.
Fig. 334
3.106. A weight W is suspended from 3 equal strings of length I which are attached to the 3 vertices
of a horizontal equilateral triangle of side s. Find the tensions in the strings.
Ans. Wl/^/91 2  3s 2
3.107. Work Problem 3.106 if there are n equal strings attached to the n vertices of a regular polygon
having n sides.
CHAP. 3] MOTION IN A UNIFORM FIELD. FALLING BODIES AND PROJECTILES
85
3.108.
3.109.
A rope passes over a fixed pulley A of Fig. 335. At one end
of this rope a mass M l is attached. At the other end of the
rope there is a pulley of mass M 2 over which passes another
rope with masses m 1 and ra 2 attached. Prove that the accel
eration of the mass m t is given by
3m 2 M 2 — m^M x — m 1 M 2 — m 2 M l — Am 1 m 2
(m x + m^Mi + M 2 ) + 4m x m 2
9
An automobile of weight W with an engine having constant
instantaneous power "P, travels up an incline of angle a.
Assuming that resistance forces are r per unit weight, prove
that the maximum speed which can be maintained up the
incline is
W(r + sin a)
S7\
KM
d
Fig. 335
3.110. An automobile of weight W moves up an incline of angle a, powered by an engine having
constant instantaneous power e P. Assuming that the resistance to motion is equal to kv per unit
weight where v is the instantaneous speed and k is a constant, prove that the maximum speed
which is possible on the incline is (y/W 2 sin 2 a + 4k WP — W sin <x)/2kW.
3.111. A chain hangs over a smooth peg with length a on one side and length 6, where < b < a, on the
U+b /v / a+V / & >
other side. Prove that the time taken for the chain to slide off is given by \ j —  — In — — —
3.112. Prove that a bead P which is placed anywhere on a vertical frictionless wire [see Fig. 336] in the
form of a cycloid
x — b(e + sin e), y = 6(1 — cos e)
will reach the bottom in the same time regardless of the starting point and find this time.
Ans. vy/b/g
y
Fig. 336
Chapter J,,
The SIMPLE HARMONIC
OSCILLATOR and the
SIMPLE PENDULUM
E
nm^
m
(a)
l + x
THE SIMPLE HARMONIC OSCILLATOR
In Fig. 4l(a) the mass m lies on a friction
less horizontal table indicated by the x axis.
It is attached to one end of a spring of negligible
mass and unstretched length I whose other end
is fixed at E.
If m is given a displacement along the x axis
[see Fig. 41(6)] and released, it will vibrate or
oscillate back and forth about the equilibrium
position 0.
To determine the equation of motion, note
that at any instant when the spring has length
I + x [Fig. 41(6)] there is a force tending to re
store m to its equilibrium position. According
to Hooke's law this force, called the restoring
force, is proportional to the stretch x and is
given by
F R =  K xi (1)
where the subscript R stands for "restoring force" and where « is the constant of propor
tionality often called the spring constant, elastic constant, stiffness factor or modulus of
elasticity and i is the unit vector in the positive x direction. By Newton's second law we have
E
nmr^
Fig. 41
d 2 (xi)
m ~aW = ~ KX1
or mx + kx
(2)
This vibrating system is called a simple harmonic oscillator or linear harmonic oscillator.
This type of motion is often called simple harmonic motion.
AMPLITUDE, PERIOD AND FREQUENCY
OF SIMPLE HARMONIC MOTION
If we solve the differential equation (2) subject to the initial conditions x = A and
dx/dt = at t = 0, we find that
(3)
x = A cos o>t where o> = s/K/m
For the case where A = 20, m = 2 and * = 8, see Problem 4.1.
Since cos tat varies between —1 and +1, the mass oscillates between x = —A and x = A.
A graph of x vs. t appears in Fig. 42.
86
CHAP. 4] THE SIMPLE HARMONIC OSCILLATOR AND THE SIMPLE PENDULUM 87
Fig. 42
The amplitude of the motion is the distance A and is the greatest distance from the
equilibrium position.
The period of the motion is the time for one complete oscillation or vibration [some
times called a cycle] such as, for example, from x = A to x = —A and then back to
x = A again. If P denotes the period, then
p = 2ttU = 2ir^/mU {U)
The frequency of the motion, denoted by /, is the number of complete oscillations or
cycles per unit time. We have
f = P = 2^ = 2^\m (5)
In the general case, the solution of (2) is
x = A cos mt + B sin o>t where w = VkTm (6)
where A and B are determined from initial conditions. As seen in Problem 4.2, we can
write (6) in the form
x = C cos (<at — <£) where <o = y/T/m (7)
and where C = y/A 2 + B 2 and <f> = tan' 1 (B/A) (8)
The amplitude in this case is C while the period and frequency remain the same as in
(4) and (5), i.e. they are unaffected by change of initial conditions. The angle <f> is called
the phase angle or epoch chosen so that ^ </> ^ v. If = 0, (7) reduces to (3).
ENERGY OF A SIMPLE HARMONIC OSCILLATOR
If T is the kinetic energy, V the potential energy and E = T + V the total energy of a
simple harmonic oscillator, then we have
T = %mv 2 , V = $kx 2 (9)
and E = \mv 2 + \ K x 2 (10)
See Problem 4.17.
THE DAMPED HARMONIC OSCILLATOR
In practice various forces may act on a harmonic oscillator, tending to reduce the
magnitude of successive oscillations about the equilibrium position. Such forces are some
times called damping forces. A useful approximate damping force is one which is propor
tional to the velocity and is given by
dx
F D = Pv = Pvi = jB^i (11)
88
THE SIMPLE HARMONIC OSCILLATOR AND THE SIMPLE PENDULUM [CHAP. 4
where the subscript D stands for "damping force" and where /? is a positive constant
called the damping coefficient. Note that F D and v are in opposite directions.
If in addition to the restoring force we assume the damping force (11), the equation
of motion of the harmonic oscillator, now called a damped harmonic oscillator, is given by
d 2 x n dx d 2 x , n dx , A
mT7z — —kx — /? — or m^ns + /3~rr + kX 
dt 2 ~ *" M dt WA "" dt 2 ' ^ dt
on applying Newton's second law. Dividing by m and calling
/?/m = 2y, kItti = <o 2
this equation can be written
X + 2y# + o> 2 X =
where the dots denote, as usual, differentiation with respect to t.
(12)
(13)
(U)
OVERDAMPED, CRITICALLY DAMPED AND
UNDERDAMPED MOTION
Three cases arise in obtaining solutions to the differential equation (14).
Case 1, Overdamped motion, y 2 > <o 2 , i.e. /? 2 > 4«ra
In this case (14) has the general solution
x = e' yt (Ae at + Be~ at ) where a = yV  «> 2 i 15 )
and where the arbitrary constants A and B can be found from the initial conditions.
Case 2, Critically damped motion, y 2 '= w 2 , i.e. p 2 = 4 K m
In this case (14) has the general solution
x = e~ yt (A + Bt) (16)
where A and B are found from initial conditions.
Case 3, Underdamped or damped oscillatory motion, y 2 < <o 2 , i.e. /3 2 < 4*<ra
In this case (14) has the general solution
x = e~ yt (A sin \t + B cos \t)
= Ce~ yt cos (\t<t>) where A = V«> 2  Y 2 i 17 )
and where C = V^ 2 + ^ 2 » called the amplitude and <£, called the phase angle or epoch,
are determined from the initial conditions.
In Cases 1 and 2 damping is so large that no
oscillation takes place and the mass m simply
returns gradually to the equilibrium position
x = 0. This is indicated in Fig. 43 where we
have assumed the initial conditions x = x ,
dx/dt — 0. Note that in the critically damped
case, mass m returns to the equilibrium position
faster than in the overdamped case.
In Case 3, damping has been reduced to such
an extent that oscillations about the equilibrium
position do take place, although the magnitude
of these oscillations tend to decrease with time
as indicated in Fig. 43. The difference in times
Critically damped motion, y* = a?
Overdamped motion, y* > «*
Underdamped motion, y 2 < u*
Fig. 43
CHAP. 4] THE SIMPLE HARMONIC OSCILLATOR AND THE SIMPLE PENDULUM 89
between two successive maxima [or minima] in the underdamped [or damped oscillatory]
motion of Fig. 43 is called the period of the motion and is given by
p = ?![ = 2tt _ 47rm , lg .
A \A> 2  y 2 \/4kWI  p 2
and the frequency, which is the reciprocal of the period, is given by
1 _ k_ V^ 7 ?" _ V^m  p 2
f ~ P ~ 2tt ~ 2tt ~ 47rm y '
Note that if p  0, (18) and (19) reduce to (4) and (5) respectively. The period and
frequency corresponding to /? = are sometimes called the natural period and natural
frequency respectively.
The period P given by (18) is also equal to two successive values of t for which
cos(A£ — <£) = 1 [or cos(Ai — <j>) = —1] as given in equation (17). Suppose that the values
of x corresponding to the two successive values t n and t n +i = t n + P are x n and x n +i respec
tively. Then
xjxn+i = e^/e^ +P) = e^ p (20)
The quantity 8 = In (xjx n +i) = yP (21)
which is a constant, is called the logarithmic decrement.
FORCED VIBRATIONS
Suppose that in addition to the restoring force — kx\ and damping force —fSvi we impress
on the mass m a force F(t)i where
F(t) = F cos at (22)
Then the differential equation of motion is
d 2 x dx
m rp = —kX  /? 7T + Fq cos at (23)
Or X + 2yi + oy 2 X = f Q COS at (2U)
where y = p/2m, w 2 = K /m, f = FJm (25)
The general solution of (24) is found by adding the general solution of
X + 2yX + o> 2 X = (26)
[which has already been found and is given by (15), (16) or (17)] to any particular solution of
(24). A particular solution of (24) is given by [see Problem 4.18]
x = ° cos (at  «/,) (27)
vV<o 2 ) 2 + 4yV
where tan d, = * ya . ^ «/» ^ tt (^5)
a — w
Now, as we have seen, the general solution of (26) approaches zero within a short time
and we thus call this solution the transient solution. After this time has elapsed, the motion
of the mass m is essentially given by (27) which is often called the steadystate solution.
The vibrations or oscillations which take place, often called forced vibrations or forced
oscillations, have a frequency which is equal to the frequency of the impressed force but
lag behind by the phase angle 4>.
90
THE SIMPLE HARMONIC OSCILLATOR AND THE SIMPLE PENDULUM [CHAP. 4
RESONANCE
The amplitude of the steadystate oscillation (27) is given by
/
oA = ° = (29)
VVa> 2 ) 2 + 4 y V
assuming y ^ 0, i.e. p ¥° 0, so that damping is assumed to be present. The maximum value
of cA in this case occurs where the frequency «/2tt of the impressed force is such that
a 2 = a 2 R = w 2 2y 2 (30)
assuming that y 2 < « 2 [see Problem 4.19]. Near this frequency very large oscillations may
occur, sometimes causing damage to the system. The phenomenon is called resonance and
the frequency ajln is called the frequency of resonance or resonant frequency.
The value of the maximum amplitude at the resonant frequency is
u
cA,
2y\A, 2  y 2
The amplitude (29) can be written in terms of a R as
aA =
VVa 2 ) 2 + 4yVy 2 )
(31)
(32)
A graph of cA vs. a 2 is shown in Fig. 44. Note that the graph is symmetric around the
resonant frequency and that the resonant frequency, frequency with damping and natural
frequency (without damping) are all different. In case there is no damping, i.e. y = or
/? = 0, all of these frequencies are identical. In such case resonance occurs where the
frequency of the impressed force equals the natural frequency of oscillation. The general
solution for this case is
x = A cos <at + B sin <at + 75— sin at (33)
From the last term in (33) it is seen that the oscillations build up with time until finally
the spring breaks. See Problem 4.20.
qA
Resonant frequency
Frequency with damping
Natural frequency
without damping)
v/////////////// (
Fig. 44
Fig. 45
THE SIMPLE PENDULUM
A simple pendulum consists of a mass m [Fig. 45] at the end of a massless string or rod
of length I [which always remains straight, i.e. rigid]. If the mass m, sometimes called the
pendulum bob, is pulled aside and released, the resulting motion will be oscillatory.
Calling the instantaneous angle which the string makes with the vertical, the
differential equation of motion is [see Problem 4.23]
CHAP. 4] THE SIMPLE HARMONIC OSCILLATOR AND THE SIMPLE PENDULUM 91
assuming no damping forces or other external forces are present.
For small angles [e.g. less than 5° with the vertical], sin 9 is very nearly equal to 0, where
9 is in radians, and equation (3U) becomes, to a high degree of approximation,
§ = f» w
This equation has the general solution
9 = A cos Vgllt + Bsin^/gllt (36)
where A and B are determined from initial conditions. For example, if 9 = 0o, = at
t = 0, then ,
9 = 6o cos Vgllt (37)
In such case, the motion of the pendulum bob is that of simple harmonic motion. The period
is given by ,
P = 2xyJTig (38)
and the frequency is given by
f = ¥ = ^^ fl {39)
If the angles are not necessarily small, we can show [see Problems 4.29 and 4.30]
that the period is equal to
iir/2 dB
P = 4
fir.
Fj = — Kjzi
VI  W sin 2 9
where k = sin (9 /2). For small angles this reduces to (38).
For cases where damping and other external forces are considered, see Problems 4.25
and 4.114.
THE TWO AND THREE DIMENSIONAL HARMONIC OSCILLATOR
Suppose a particle of mass m moves in the xy plane y
under the influence of a force field F given by
F =  Kl a?i  K 2 yj (U)
where k x and k 2 are positive constants.
In this case the equations of motion of m are
given by j
m dP = ~ K i x ' m d& = ~ K * y ^
and have solutions Fig. 46
x = A 1 cos y/ajm t + B x sin yj«.jm t, y = A 2 cos yj*.jm t + B 2 sin y/ajm t (A3)
where Ai, B\, A%, B 2 are constants to be determined from the initial conditions. The mass m
subjected to the force field (41) is often called a twodimensional harmonic oscillator. The
various curves which m describes in its motion are often called Lissajous curves or figures.
These ideas are easily extended to a three dimensional harmonic oscillator of mass m
which is subject to a force field given by
F = k x x\  K 2 y]  K 3 zk (U)
where < lf k 2 , « 3 are positive constants.
m
F 2 = /c 2 l/j
92 THE SIMPLE HARMONIC OSCILLATOR AND THE SIMPLE PENDULUM [CHAP. 4
Solved Problems
SIMPLE HARMONIC MOTION AND THE
SIMPLE HARMONIC OSCILLATOR
4.1. A particle P of mass 2 moves along the x axis attracted toward origin O by a force
whose magnitude is numerically equal to Sx [see Fig. 47]. If it is initially at rest
at x = 20, find (a) the differential equation and initial conditions describing the
motion, (b) the position of the particle at any time,
(c) the speed and velocity of the particle at any time,
and (d) the amplitude, period and frequency of the
vibration.
8a?i
xi
(a) Let r = xi be the position vector of P. The acceleration {
d 2 d 2 x
of P is jijjC^i) ~ ~d+2*' ^ ne ne t f° rce acting on P is
— 8xi. Then by Newton's second law, Fig. 47
n d 2 x. „ . d 2 x , . .„.
2^1 = 8jbi or ^p+ 4x = (1)
which is the required differential equation of motion. The initial conditions are
x = 20, dx/dt = at t = (2)
(6) The general solution of (1) is
x = A cos It + B sin 2£ (3)
When t = 0, x = 20 so that A = 20. Thus
x = 20 cos It + B sin 2t (4)
Then dx/dt = 40 sin 2t + 2B cos 2t (5)
so that on putting t = 0, dx/dt = we find B = 0. Thus (5) becomes
a; = 20 cos 2i (0)
which gives the position at any time.
(c) From (6) dx/dt = 40 sin It which gives the speed at any time. The velocity is given by
^i = 40sin2ti
dt
(d) Amplitude = 20. Period = 2tt/2 = w. Frequency = 1/period = 1/v.
4.2. (a) Show that the function A cos at + B sin «>t can be written as C cos (U  </>)
where C = V A 2 + B 2 and <j> = tan 1 (S/A). (6) Find the amplitude, period and
frequency of the function in (a).
(a) A cos at + B sin at = \l A 2 + B 2 ( cos at + sin w A
W^ 2 + B 2 VA 2 + B 2 /
= \/A 2 + # 2 (cos <(> cos w£ + sin <f> sin to*)
= VA 2 + B 2 cos («t  *) = Ccos(wt0)
where cos = A/ VA 2 + £ 2 and sin = B/y/A 2 + B 2 , i.e. tan <p = 5/ A or = tan 1 £/A,
and C = VA 2 + £ 2 . We generally choose that value of <p which lies between 0° and 180°,
i.e. g <p ^ v.
(b) Amplitude = maximum value = C  y/A^TB 2 . Period = 2a/«. Frequency = w/2*.
CHAP. 4] THE SIMPLE HARMONIC OSCILLATOR AND THE SIMPLE PENDULUM 93
4.3. Work Problem 4.1 if P is initially at x = 20 but is moving (a) to the right with speed
30, (b) to the left with speed 30. Find the amplitude, period and frequency in
each case.
(a) The only difference here is that the condition dx/dt = at * = of Problem 4.1 is replaced
by dx/dt = 30 at t ~ 0. Then from (5) of Problem 4.1 we find B = 15, and (3) of
Problem 4.1 becomes
x = 20 cos 2t + 15 sin 2t CO
\
which gives the position of P at any time. This may be written [see Problem 4.2] as
x = V(20) 2 + (15)2 I 20 — cos 2t + 15 ; sin 2t \
1/(20)2+ (15) 2 V(20)2 + (15)2 J
= 25{cos2i + f sin2£} = 25 cos (2t  0)
where cos = f, sin = f (2)
The angle which can be found from (2) is often called the phase angle or epoch.
Since the cosine varies between —1 and +1, the amplitude = 25. The period and fre
quency are the same as before, i.e. period = 2w/2 = w and frequency = 2/2?r = II v.
(b) In this case the condition dx/dt = at t = of Problem 4.1 is replaced by dx/dt = 30
at t = 0. Then B — —15 and the position is given by
x = 20 cos It — 15 sin 2t
which as in part (a) can be written
x = 25{cos2t f sin2t}
= 25{cos cos 2t + sin ^ sin 2i) = 25 cos (2t — 0)
where cos f = f , sin ^ = — f .
The amplitude, period and frequency are the same as in part (a). The only difference
is in the phase angle. The relationship between ^ and <f> is $ = <p + w. We often describe
this by saying that the two motions are 180° out of phase with each other.
4.4. A spring of negligible mass, suspended vertically from one end, is stretched a
distance of 20 cm when a 5 gm mass is attached to the other end. The spring and
mass are placed on a horizontal frictionless table as in Fig. 4l(a), page 86, with the
suspension point fixed at E. The mass is pulled away a distance 20 cm beyond the
equilibrium position O and released. Find (a) the differential equation and initial
conditions describing the motion, (b) the position at any time t, and (c) the amplitude,
period and frequency of the vibrations.
(a) The gravitational force on a 5 gm mass [i.e. the weight of a 5 gm mass] is 5g = 5(980) dynes =
4900 dynes. Then since 4900 dynes stretches the spring 20 cm, the spring constant is
K = 4900/20 = 245 dynes/cm. Thus when the spring is stretched a distance x cm beyond the
equilibrium position, the restoring force is — 245xi. Then by Newton's second law we have,
if r = xi is the position vector of the mass,
5 *g*> = _245zi or ^+ 49* = (1)
The initial conditions are x = 20, dx/dt = at t = (2)
(b) The general solution of (1) is x = A cos It + B sin It ($)
Using the conditions (*) we find A = 20, B = so that x  20 cos It.
(c) From x = 20 cos It we see that: amplitude = 20 cm; period = 2tt/7 sec; frequency = 7/2ir
vib/sec or l/2v cycles/sec.
94
THE SIMPLE HARMONIC OSCILLATOR AND THE SIMPLE PENDULUM [CHAP. 4
4.5. A particle of mass m moves along the x axis, attracted toward a fixed point O
on it by a force proportional to the distance from O. Initially the particle is at
distance x from and is given a velocity v away from 0. Determine (a) the
position at any time, (b) the velocity at any time, and (c) the amplitude, period,
frequency, and maximum speed.
(a) The force of attraction toward O is — kx'i where k is a
positive constant of proportionality. Then by Newton's
second law,
d?x.
•• . KX
—kxi or x H =0
m
Solving (1), we find
x = A cos y/ic/m t + B sin yjn/m t
We also have the initial conditions
(1)
(2)
F = — Kxi
m
Fig. 48
x = x , dx/dt — v at t =
From x = x at t = we find, using (2), that A = x . Thus
x = x cos V ' ii/m t + B sin yic/m t
so that dx/dt = —x Q yjulm sin V*/™ t + B y/lc/m cos yV™ £
From dx/dt = v at t = we find, using (5), that B = v vWk. Thus (4) becomes
a; = #0 cos V~i</ni t + v Q \Zm/K sin vk/w t
Using Problem 4.2, this can be written
where
x = yx^+mvUic cos (vk/ot £ — 0)
^ = tan 1 (v /x ) yW/c
(3)
(•4)
(5)
(8)
(7)
(8)
(6) The velocity is, using (0) or (7),
dx .
v = tt i = (— #o y/x/m sin VxTm £ + v cos yVm £) i
= —^Jit/m \/x 2 + mv 2 /ic sin (yfic/mt — <j>) i
= — V^n + k«?/w. sin {\[7hn t — 0) i
(3)
(c) The amplitude is given from (7) by yxJ+mv*/K.
From (7), the period is P = 2srV 'ic/m. The frequency is / = 1/P = 2iry/m/K.
From (3), the speed is a maximum when sin (y/T/mt </>) = ±1; this speed is v^ 2 , + kb 2 /™.
4.6. An object of mass 20 kg moves with simple harmonic motion on the x axis. Initially
(t = 0) it is located at the distance 4 meters away from the origin x = 0, and has
velocity 15 m/sec and acceleration 100 m/sec 2 directed toward * = 0. Find (a) the
position at any time, (b) the amplitude, period and frequency of the oscillations, and
(c) the force on the object when t  rr/10 sec.
(a) If x denotes the position of the object at time t, then the initial conditions are
x = 4, dx/dt = 15, d 2 x/dt 2 = 100 at t =
Now for simple harmonic motion,
= A cos u>t + B sin at
Differentiating, we find
dx/dt = —Aw sin at + Ba cos at
d 2 x/dt 2 = —Aw 2 cos ut  Ba 2 sin at
(1)
(3)
(4)
CHAP. 4] THE SIMPLE HARMONIC OSCILLATOR AND THE SIMPLE PENDULUM
95
Using conditions (1) in {2), (S) and (4), we find 4 = A, 15 = Bw, 100 = Aw 2 . Solving
simultaneously, we find A = 4, w = 5, 5 = — 3 so that
x = 4 cos 5t — 3 sin 5t
which can be written
x = 5 cos (5t — #) where cos ^ = ^ , sin ^ = — §
(6) From (tf) we see that: amplitude = 5 m, period = 2jt/5 sec, frequency = 5/2jr vib/sec.
(c) Magnitude of acceleration = d 2 x/dt 2 = —100 cos 5t + 75 sin 5* = 75 m/sec 2 at t = tt/10.
Force on object = (mass) (acceleration) = (20 kg)(75 m/sec 2 ) = 1500 newtons.
(5)
(6)
4.7.
4.8.
A 20 lbwt object suspended from the end of a ver
tical spring of negligible mass stretches it 6 inches.
(a) Determine the position of the object at any time
if initially it is pulled down 2 inches and then re
leased, (b) Find the amplitude, period and fre
quency of the motion.
(a) Let D and E [Fig. 49] represent the position of the
end of the spring before and after the object is put on
the spring. Position E is the equilibrium position of
the object.
Choose a coordinate system as shown in Fig. 49
so that the positive z axis is downward with origin at
the equilibrium position.
By Hooke's law, since 20 lb wt stretches the spring
\ ft, 40 lb wt stretches it 1 ft; then 40(.5 + z) lb wt
stretches it (.5 + z) ft. Thus when the object is at
position F there is an upward force acting on it of
magnitude 40(.5 + z) and a downward force due to its
weight of magnitude 20. By Newton's second law we
thus have
H^k = 20k  40(.5 + z)k or
mm^x?
D
E
J .5 ft
zft
F
F 1
I I
Fig. 49
W2 + 64, =
Solving, z = A cos St + B sin St
Now at t = 0, z = % and dz/dt = 0; thus A = £, B = and
(1)
z = ^ cos 8£
(b) From (2): amplitude = £ ft, period = 2jt/8 = v/4 sec, frequency = 4/v vib/sec.
Work Problem 4.7 if initially the object is pulled down 3 inches (instead of 2 inches)
and then given an initial velocity of 2 ft/sec downward.
In this case the solution (1) of Problem 4.7 still holds but the initial conditions are: at t = 0,
z — \ and dz/dt = 2. From these we find A = \ and B = \, so that
z = I cos St + I sin Bt = y/2/4 cos (St  ir/4)
Thus amplitude = \/2/4 ft, period = 2a/8 = jt/4 sec, frequency = 4Ar vib/sec. Note that the period
and frequency are unaffected by changing the initial conditions.
4.9. A particle travels with uniform angular speed a around a circle of radius 6. Prove
that its projection on a diameter oscillates with simple harmonic motion of period
2tt/g> about the center.
Choose the circle in the xy plane with center at the origin O as in Fig. 410 below. Let Q be
the projection of particle P on diameter AB chosen along the x axis.
96
THE SIMPLE HARMONIC OSCILLATOR AND THE SIMPLE PENDULUM [CHAP. 4
If the particle is initially at B, then in time t we will
have I BOP — e — ut. Then the position of P at time t is
r = b cos ut i + b sin ut j
The projection Q of P on the x axis is at distance
r • i — x = b cos ut
(1)
(2)
from O at any time t. From (2) we see that the projection Q
oscillates with simple harmonic motion of period 2ir/w about
the center O.
Fig. 410
THE DAMPED HARMONIC OSCILLATOR
4.10. Suppose that in Problem 4.1 the particle P has also a damping force whose magnitude
is numerically equal to 8 times the instantaneous speed. Find (a) the position and
(b) the velocity of the particle at any time, (c) Illustrate graphically the position of
the particle as a function of time t. y
(a)
(b)
(c)
In this case the net force acting on P is [see
doc
— 8a;i — 8rri. Then by Newton's sec
Fig. 411]
ond law,
2— i
dt
= 8«i
d 2 x
+ ^ + i *
, dx
di
 8^i
1
Fig. 411
P
This has the solution [see Appendix, page 352, Problem C.14]
x = e»(A + Bt)
When t = 0, x  20 and dx/dt = 0; thus A = 20, B = 40, and a; = 20e~ 2t (1 + 2t) gives
the position at any time t.
The velocity is given by
dx .
V = ^
80«e«i
The graph of a; vs. £ is shown in Fig. 412. It is
seen that the motion is nonoscillatory. The par
ticle approaches O slowly but never reaches it.
This is an example where the motion is critically
damped.
Fig. 412
4.11. A particle of mass 5 gm moves along the x axis under the influence of two forces:
(i) a force of attraction to origin O which in dynes is numerically equal to 40 times
the instantaneous distance from O, and (ii) a damping force proportional to the
instantaneous speed such that when the speed is 10 cm/sec the damping force is
200 dynes. Assuming that the particle starts from rest at a distance 20 cm from O,
(a) set up the differential equation and conditions describing the motion, (b) find
the position of the particle at any time, (c) determine the amplitude, period and
frequency of the damped oscillations, and (d) graph the motion.
(a) Let the position vector of the particle P be denoted by
r = xi as indicated in Fig. 413. Then the force of attrac
tion (directed toward O) is
40»i (1)
The magnitude of the damping force / is proportional to
the speed, so that / = (S dx/dt where /? is constant. Then
since / = 200 when dx/dt = 10, we have /? = 20 and
/ = 20 dx/dt. To get £, note that when dx/dt > and
x > the particle is on the positive x axis and moving to
O
20(dx/dt)i
— 40*i
Fig. 413
CHAP. 41 THE SIMPLE HARMONIC OSCILLATOR AND THE SIMPLE PENDULUM
97
the right. Thus the resistance force must be directed toward the left. This can only be accom
plished if ^
f =  20 ^
(2)
This same form for f is easily shown to be correct if * > 0, dx/dt < 0, x < 0, dx/dt > 0,
x < 0, dx/dt < [see Problem 4.45].
Hence by Newton's second law we have
d*x.
b dP 1
20%  40*1
dt
d z x , . dx .
1M + *M + SX
(3)
04)
(5)
Since the particle starts from rest at 20 cm from O, we have
* = 20, dx/dt = at * =
where we have assumed that the particle starts on the positive side of the x axis [we could
just as well assume that the particle starts on the negative side, in which case x — —20].
(b) x = e at is a solution of (4) if
a 2 + 4a + 8 =
Then the general solution is
i(4 ± V16  32) = 2 ± 2%
x = e~ 2t (A cos2t + B sin2t) (6)
Since x = 20 at t = 0, we find from (6) that A = 20, i.e.,
x = e~ 2t (20 cos 2t + B sin 2t) (7)
Thus by differentiation,
dx/dt = (e 2 t)(40 sin 2* + 2B cos 2t) + (2e~2t)(20 cos 2t + B sin 2t) (8)
Since dx/dt = at t = 0, we have from (8), B = 20. Thus from (7) we obtain
a; = 20e2'(cos 2t + sin 2t) = 20\/2 e" 2 * cos (2t  ir/4) (9)
using Problem 4.2.
(c) From (9): amplitude = 20V2 e _2t cm, period = 2jt/2 = tt sec, frequency = 1/V vib/sec.
(d) The graph is shown in Fig. 414. Note that the amplitudes of the oscillation decrease toward
zero as t increases.
20V2
Fig. 414
4.12. Find the logarithmic decrement in Problem 411.
Method 1. The maxima (or minima) of * occur where dx/dt = 0. From (9) of Problem 4.11,
dx/dt  80e 2t sin2£ =
when t — 0, ir/2, w, 3n/2, 2ir, 5ir/2, .... The maxima occur when t = 0, v, 2v, . . . ; the minima
occur when t = n/2, 3s/2, 5xr/2, .... The ratio of two successive maxima is e 2 (0)/ e 2(7r) or
e 2(ir)/ e 2(27r) > etc., i.e. e 2v . Then the logarithmic decrement is 8 =ln(e 2w ) = 2w.
98 THE SIMPLE HARMONIC OSCILLATOR AND THE SIMPLE PENDULUM [CHAP. 4
Method 2.
From (9) of Problem 4.11, the difference between two successive values of t, denoted by t n
and t n + 1 , for which cos(2t — tt/4) — 1 (or —1) is w, which is the period. Then
Xn 20\/2e2tn
— — = — = e 27r and 8 — \n(x n /x n + 1 ) = 2b
x n + i 20y/2e 2t n+i
Method 3. From (13), (18) and (21), pages 88 and 89, we have
8 = yP
— ( £ \( ^ irm \ = 2 ^ft
Then since ra = 5, p — 20, k = 40 [Problem 4.11, equation (3)], 8 = 2w.
4.13. Determine the natural period and frequency of the particle of Problem 4.11.
The natural period is the period when there is no damping. In such case the motion is
given by removing the term involving dx/dt in equation (3) or (4) of Problem 4.11. Thus
d 2 x/dt 2 + Sx = or a; = A cos 2y[2t + B sin 2y/2t
Then: natural period = 2irl2y[2 sec = irly[2 sec; natural frequency = y[2h vib/sec.
4.14. For what range of values of the damping constant in Problem 4.11 will the motion
be (a) overdamped, (b) underdamped or damped oscillatory, (c) critically damped?
Denoting the damping constant by p, equation (3) of Problem 4.11 is replaced by
p d 2 x. n dx . .. . d 2 x , P dx _ _ A
5 dtff = ** 1  40 * 1 ° r d# + 5Tt +8x ' °
Then the motion is:
(a) Overdamped if (/3/5) 2 > 32, i.e. p > 20 v / 2.
(b) Underdamped if (p/5) 2 < 32, i.e. p < 20^2.
[Note that this is the case for Problem 4.11 where p = 20.]
(c) Critically damped if (p/5) 2 = 32, i.e. p = 20V2.
4.15. Solve Problem 4.7 taking into account an external damping force given numerically
in lb wt by pv where v is the instantaneous speed in ft/sec and (a) /? = 8, (b) p = 10,
(c) p = 12.5.
The equation of motion is
20 d 2 z „ dz. d 2 z , SB dz , _. .
I^ k = 20k40(.5 + ,)k/ 3 k or ^ + f ^ + 64, =
(a) If /8 = 8, then d 2 z/dt 2 + 12.8 dz/dt + 642 = 0. The solution is
z = e~ 6At (A cos4.8£ + J? sin 4.8t)
Using the conditions z = 1/6, dz/dt = at * = 0, we find A = 1/6, J3 = 2/9 so that
2 = ?e6.4t(3 C0S 4.8£ + 4 sin 4.8*) = Jg e ~ 8 ' 4t cos < 4  8t ~ 53 ° 8 '>
The motion is damped oscillatory with period 2?r/4.8 = 5V/12 sec.
(6) If p = 10, then d 2 z/dt 2 + Udz/dt + 64z = 0. The solution is
z = e~ 4t (A+Bt)
Solving subject to the initial conditions gives A = %, B = f ; then %  \e~ 4t (1 + U).
CHAP. 4] THE SIMPLE HARMONIC OSCILLATOR AND THE SIMPLE PENDULUM
99
The motion is critically damped since any decrease in /? would produce oscillatory motion.
(c) If /? = 12.5 then d 2 z/dt 2 + 20dz/dt + 64z = 0. The solution is
z = Aeu + Be™
Solving subject to initial conditions gives A = 1/6, B = 1/24; then z  \e~^  ^e~ 16t .
The motion is overdamped.
ENERGY OF A SIMPLE HARMONIC OSCILLATOR
4.16. (a) Prove that the force F =  K xi acting on a simple harmonic oscillator is con
servative, (b) Find the potential energy of a simple harmonic oscillator.
(a) We have V X F
i J k
d/dx d/dy d/dz
—kx
= so that F is conservative.
(6) The potential or potential energy is given by V where FVV or
fdV. ,BV. ,8V.
~ KXl = (jr + ^ J + *T k
Then dV/dx = kx, dV/dy = 0, dV/dz = from which V = fax 2 + c. Assuming V = cor
responding to x = 0, we find c = so that V = fax 2 .
4.17. Express in symbols the principle of conservation of energy for a simple harmonic
oscillator.
By Problem 4.16(6), we have
Kinetic energy + Potential energy = Total energy
or fafiv 2 + fax 2 = E
which can also be written, since v = dx/dt, as fan{dx/dt) 2 + fax 2 = E.
Another method. The differential equation for the motion of a simple harmonic oscillator is
md 2 x/dt 2 = — kx
Since dx/dt — v, this can also be written as
m li = ~ KX ° r m dxlt = ~ KX >
dv dv dx
,— = — K x or
dt
Integration yields fanv 2 + fax 2 = E.
dv
i.e. mv= = —kx
dx
FORCED VIBRATIONS AND RESONANCE
4.18. Derive the steadystate solution (27) corresponding to the differential equation (24)
on page 89.
The differential equation is ••,„•, *
x + 2ya? + u 2 x = / cos at
Consider a particular solution having the form
X = Ci cos at + c 2 sin at
(1)
(*)
100 THE SIMPLE HARMONIC OSCILLATOR AND THE SIMPLE PENDULUM [CHAP. 4
where c x and c 2 are to be determined. Substituting (2) into (1), we find
(a 2 *?! + 2yaC 2 + u 2 c t ) cos at + (—a 2 c 2 — 2yac 1 + w 2 c 2 ) sin at = / cos at
from which « 2 Cl + 2yac 2 + w 2 Cl = / ,  a 2 c 2  2yac x + « 2 c 2 = (3)
or (a 2 ~ w 2 )cj  2yac 2 = /„, 2yaC x + (a 2  <* 2 )c 2 = (4)
Solving these simultaneously, we find
— /o (<o 2 — a 2 ) _ 2/ yw
Cl ~ (a 2  <o 2 ) 2 + 4y 2 <o 2 ' C2 ~ (a 2  <o 2 ) 2 + 4y 2 a 2 (5)
Thus (2) becomes
_ /o [(" 2 — « 2 ) cos at + 2ya sin at]
* ~ (a 2  co 2 ) 2 + 4y 2 a 2 ( e )
Now by Problem 4.2, page 92,
(<o 2  a 2 ) cos at + 2ya sin at  V(« 2 ~ a 2 ) 2 + 4y 2 a 2 cos (at  <f>) (7)
where tan <p = 2ya/(a 2  <o 2 ), ^ <j> ^ v. Using (7) in (5), we find as required
x = — cos (at — 0)
V(« 2 ~ <o 2 ) 2 + 4y 2 a 2
4.19. Prove (a) that the amplitude in Problem 4.18 is a maximum where the resonant fre
quency is determined from a = \/«> 2  2y 2 and (b) that the value of this maximum
amplitude is / /(2y\/ w 2 y 2 ).
Method 1. The amplitude in Problem 4.18 is
/o/vV  <o 2 ) 2 + 4y 2 « 2 (1)
It is a maximum when the denominator [or the square of the denominator] is a minimum. To
find this minimum, write
(a 2  <o 2 ) 2 + 4y 2 a 2 = a 4  2(<o 2  2y 2 )a 2 + w 4
= a 4  2(w 2  2y 2 )a 2 + (a> 2  2y 2 ) 2 + <o 4  (<o 2  2y 2 ) 2
= [a 2  (« 2  2y 2 )] 2 + 4y 2 (a> 2  y 2 )
This is a minimum where the first term on the last line is zero, i.e. when a 2 = w 2 — 2y 2 , and the
value is then 4y 2 (w 2 — y 2 ). Thus the value of the maximum amplitude is given from (i) by
/o/^yV^Y 2 )
Method 2. The function U — (a 2 — w 2 ) 2 + 4y 2 a 2 has a minimum or maximum when
$T = 2(a 2  a> 2 )2a + 8y 2 a = or a(a 2  <o 2 + 2y 2 ) =
da
i.e. a = 0, a — Vw 2 — 2y 2 where y 2 < ^w 2 . Now
dPU/da 2 = 12a 2  4<o 2 + 8y 2
For « = 0, d 2 U/da 2 = 4(<o 2  2y 2 ) < 0. For a = Vw 2  2y 2 , d 2 U/da 2 = 8(a> 2  2y 2 ) > 0. Thus
a = V" 2 — 2y 2 gives the minimum value.
4.20. (a) Obtain the solution (33), page 90, for the case where there is no damping and the
impressed frequency is equal to the natural frequency of the oscillation, (b) Give a
physical interpretation.
(a) The case to be considered is obtained by putting y = or (3 = and a = a in equations
{23) or (24), page 89. We thus must solve the equation
x + a 2 x = / cos tot (■*)
CHAP. 41 THE SIMPLE HARMONIC OSCILLATOR AND THE SIMPLE PENDULUM
101
To find the general solution of this equation we add the general solution of
x + u 2 x = (*)
to a particular solution of (1).
Now the general solution of (2) is
x — A cos ut + B sin ut (#)
To find a particular solution of (1) it would do no good to assume a particular solution of the form
x = c t cos ut + c 2 sin ut w
since when we substitute (4) [which is identical in form to {3)] into the left side of {1), we would
get zero. We must therefore modify the form of the assumed particular solution (4). As seen
in Appendix C, the assumed particular solution has the form
x = t{Ci cos ut + c 2 sin ut) w)
To see that this yields the required particular solution, let us differentiate (5) to obtain
x — t(— «c x sin ut + uc 2 cos ut) + (c x cos ut + c 2 sin ut)
x = t(— u 2 c x cos ut — u 2 c 2 sin ut) + 2(— uc t sin ut + uc 2 cos ut)
Substituting (5), (6) and (7) into (1), we find after simplifying
— 2uc t sin ut + 2uc 2 cos ut = f cos ut
from which c x = and c 2 = f /2u. Thus the re
quired particular solution (5) is x — (f /2u)t sin ut.
The general solution of (1) is therefore
x = A cos wt + B sin <ot + {fJ2u)t sin tot (5)
(6) The constants A and 5 in (8) are determined
from the initial conditions. Unlike the case with
damping, the terms involving A and B do not become
small with time. However, the last term involving t
increases with time to such an extent that the spring
will finally break. A graph of the last term shown
in Fig. 415 indicates how the oscillations build up
in magnitude.
(6)
(7)
Fig. 415
4.21. A vertical spring has a stiffness factor equal to 3 lb wt per ft. At £ = a force
given in lb wt by F(t) = 12 sin4£, t ^ is applied to a 6 lb weight which hangs in
equilibrium at the end of the spring. Neglecting damping, find the position of the
weight at any later time t.
Using the method of Problem 4.7, we have by Newton's second law,
6 d 2 z
32 dt 2
d?z
= Sz + 12 sin At
^ + 162 = 64 sin 4t
(1)
Solving ' z = A cos At + B sin 4t  8t cos At
When t = 0, z = and dz/dt = 0; then A = 0, B = 2 and
z  2 sin At  8t cos At (2)
As t gets larger the term —St cos At increases numerically without bound, and physically the
spring will ultimately break. The example illustrates the phenomenon of resonance. Note that
the natural frequency of the spring (4/2*r = 2/w) equals the frequency of the impressed force.
102
THE SIMPLE HARMONIC OSCILLATOR AND THE SIMPLE PENDULUM [CHAP. 4
4.22. Work Problem 4.21 if F(t) = 30 cos Gt, t^O.
In this case the equation (1) of Problem 4.21 becomes
d 2 z/dt 2 + 16z = 160 cos 5t (1)
and the initial conditions are
z = 0, dz/dt  at t  (2)
The general solution of (1) is
z  A cos4£ + B sin 4* — 8 cos 6t (8)
Using conditions (2) in (5), we find A = 8, £ = and
2 = 8(cos 4t  cos 6t) = 8{cos(5««)  cos(5£ + £)} = 16 sin t sin 5t
The graph of z vs. t is shown by the heavy curve of Fig. 416. The dashed curves are the curves
z = ±16 sin t obtained by placing sin 5t = ±1. If we consider that 16 sin t is the amplitude of
sin5«, we see that the amplitude varies sinusoidally. The phenomenon is known as amplitude
modulation and is of practical importance in communications and electronics.
Fig. 416
THE SIMPLE PENDULUM
4.23. Determine the motion of a simple pendulum of length I and mass m assuming small
vibrations and no resisting forces.
Let the position of m at any time be determined by s,
the arclength measured from the equilibrium position
[see Fig. 417]. Let o be the angle made by the pendulum
string with the vertical.
If T is a unit tangent vector to the circular path of
the pendulum bob m, then by Newton's second law
m dP T =
mg sin e T
(1)
or, since s = le,
<&8 _ g .
For small vibrations we can replace sin e by B so that
to a high degree of accuracy equation (2) can be replaced by
dH
dP
+ 5< = °
(3)
which has solution
e = A coay/gTlt + B sinVFT^*
Taking as initial conditions $ = $ , de/dt = at t = 0, we find A =
e = e cos VoU t
From this we see that the period of the pendulum is 2iry/l/g.
CHAP. 4] THE SIMPLE HARMONIC OSCILLATOR AND THE SIMPLE PENDULUM 103
4.24. Show how to obtain the equation (2) for the pendulum of Problem 4.23 by using
the principle of conservation of energy.
We see from Fig. 417 that OA = OC  AC = I  I cos e  1(1  cos o). Then by the conserva
tion of energy [taking the reference level for the potential energy as a horizontal plane through
the lowest point O] we have
Potential energy at B + Kinetic energy at B = Total energy = E = constant
mgl(l ~ cos e) + ^m(ds/dt) 2 = E (1)
Since s — Id, this becomes
mgl{\  cos e) + \ml 2 (de/dt) 2 = E (2)
Differentiating both sides of (2) with respect to t, we find
mgl sin e h + ml 2 = or V + (g/l) sin —
in agreement with equation (2) of Problem 4.23.
4.25. Work Problem 4.23 if a damping force proportional to the instantaneous velocity is
taken into account.
In this case the equation of motion (1) of Problem 4.23 is replaced by
d 2 s ds d 2 s B ds
m d^ T = mgsmeT  B Tt T or w = g sin e   ^
Using s = le and replacing sin o hy d for small vibrations, this becomes
dt* m dt I
Three cases arise:
Case 1. p 2 /4m 2 < g/l
e  e~^ t/2m (A cosut + B smut) where w = y/g/l  B 2 lkm 2
This is the case of damped oscillations or underdamped motion.
Case 2. B 2 /4m 2 = g/l
e = e~^ t/2m (A+Bt)
This is the case of critically damped motion.
Case 3. B 2 /4m 2 > g/l
e = e 3t/2m(^ e \t + J5 e xt) where X = V)8 2 /4m 2  g/l
This is the case of overdamped motion.
In each case the constants A and B can be determined from the initial conditions. In Case 1
there are continually decreasing oscillations. In Cases 2 and 3 the pendulum bob gradually returns
to the equilibrium position without oscillation.
THE TWO AND THREE DIMENSIONAL HARMONIC OSCILLATOR
4.26. Find the potential energy for (a) the two dimensional and (b) the three dimensional
harmonic oscillator.
(a) In this case the force is given by
F = — k^xi — K 2 yj
Since VXF = 0, the force field is conservative. Thus a potential does exist, i.e. there exists
a function V such that F = — VV. We thus have
dV . dV . av,
104 THE SIMPLE HARMONIC OSCILLATOR AND THE SIMPLE PENDULUM [CHAP. 4
from which dV/dx = Kl x, dV/dy  K 2 y, dV/dz = or
V = 1^*2 + ^ K2 y2
choosing the arbitrary additive constant to be zero. This is the required potential energy.
(b) In this case we have F = — k x xi — K 2 y]  K S zk which is also conservative since V X F = 0.
We then find as in part (a), dV/dx  k x x, 8V/dy = K 2 y, dV/dz = k 3 z from which the required
potential energy is
V = fax* + 1k 2 2/2 + fa z 2
4.27. A particle moves in the xy plane in a force field given by F =  K xi  K yj. Prove that
in general it will move in an elliptical path.
If the particle has mass ra, its equation of motion is
mjp = F = kxi  K yj (l)
d oc ct  i/
or, since r = xi + yj, m^i + m~j =  K xi  K yj
ct ! ic d^ij
Then m — =  KX , m^ =  K y (2)
These equations have solutions given respectively by
x = A x cos yj kItyi t + A 2 sin V^M t, y — B t cos y/ n/m t + B 2 sin V '/c/m t (3)
Let us suppose that at t — the particle is located at the point whose position vector is
r = ai + bj and movin g with ve locity dr/dt = v x i + v 2 }. Using these conditions, we find A t = a,
B t = b, A 2 — Vis/mU, B 2 = v 2 y/m/K and so
x — a cos wt + c sin wt, y = b cos ut + d sin at (4)
where c = v{\fmTic, d = v 2 y/m/K. Solving for sin ut and cosut in (4) we find, if ad ¥= be,
dx — cy . , ay — bx
cos ut = —3 — r , sin ut = — j — j—
ad — be ad— be
Squaring and adding, using the fact that cos 2 ut + sin 2 w £ = 1, we find
(dx  cy)* + (ay  bx) 2 = (ad be) 2
or (62 + d 2 )x 2  2(cd + ab)xy + (a* + c 2 )y* = (ad  be) 2 (5)
Now the equation
Ax 2 + Bxy + Cy 2 = D where A>0, C > 0, D >
is an ellipse if B 2 4AC < 0, a parabola if B 2 — 4AC = 0, and a hyperbola if B 2 4AC>0.
To determine what (5) is, we see that A = b 2 + d 2 , B = 2(cd + ab), C = a 2 + c 2 so that
B 2  4AC = 4(cd + ab) 2  4(b 2 + d?)(a 2 + c 2 ) = 4(ad  be) 2 <
provided ad ¥= be. Thus in general the path is an ellipse, and if A  C it is a circle. If ad = be
the ellipse reduces to the straight line ay = bx.
MISCELLANEOUS PROBLEMS
4.28. A cylinder having axis vertical floats in a liquid of density a. It is pushed down
slightly and released. Find the period of the oscillation if the cylinder has weight W
and cross sectional area A.
CHAP. 4] THE SIMPLE HARMONIC OSCILLATOR AND THE SIMPLE PENDULUM
105
Let RS, the equilibrium position of the cylinder, be distant
z from the liquid surface PQ at any time t. By Archimedes'
principle, the buoyant force on the cylinder is (Az)a. Then by
Newton's second law,
W d?z
9 d&
rrs" = —Azo
d?z , gAo
dt* W z
=
Solving,
z = c t cos yJgAolW t + c 2 sin ^JgAaJW t
and the period of the oscillation is 2vyW/gAo.
Fig. 418
4.29. Show that if the assumption of small vibrations is not made, then the period of a
simple pendulum is
41 S.
d<f>
VI ~k 2 sin 2 */,
where k = sin (<9 /2)
The equation of motion for a simple pendulum if small vibrations are not assumed is
[equation (34), page 91]
d 2 e a .
Let de/dt = u. Then
and (1) becomes
Integrating (2) we obtain
dt*
dH
d&
_ du
~ dt
=
du de _
de dt ~
du
de
du
de
=
fsin*
(1)
r — y cos + c
Now when e = e , u = so that c = ~(g/l) cos e . Thus (3) can be written
u 2 = (2g/l)(cos e  cos e ) or de/dt = ±^(2g/l)(cose  cos0 o )
(S)
U)
If we restrict ourselves to that part of the motion where the bob goes from e = e to = 0,
which represents a time equal to one fourth of the period, then we must use the minus sign in (4)
so that it becomes
de/dt =  y/(2g/l)(cos o — cos e )
Separating the variables and integrating, we have
• ■ us
de
Vcos e — cos e
Since t = at e = e and * = P/4 at e = 0, where P is the period,
de
[T C e °
Vcos e — cos e
(5)
Making use of the trigonometric identity cos e = 2 sin 2 (*/2) — 1, with a similar one replacing
e by e , (5) can be written
P = 2
vir
d*
Now let
Vsin 2 (0 o /2)  sin 2 (<?/2)
sin (e/2) = sin (<? /2) sin
(7)
106 THE SIMPLE HARMONIC OSCILLATOR AND THE SIMPLE PENDULUM [CHAP. 4
Then taking the differential of both sides,
\ cos (0/2) de — sin (e /2) cos <p d<f>
or calling k — sin (e /2),
2 sin (e /2) cos <p d<f>
de — —
Vl  k 2 sin 2 <f>
Also from (7) we see that when e = 0, <f> = 0; and when e = e , <j> = jt/2. Hence (6) becomes,
as required,
_ ,7r/2 d±_
\ Vl  k 2 sin 2
*  <VJr=2=
Note that if we have small vibrations, i.e. if k is equal to zero very nearly, then the period (8)
becomes
p = 4 VJC^ = ^VI
(9)
as we have already seen.
The integral in (8) is called an elliptic integral and cannot be evaluated exactly in terms of
elementary functions. The equation of motion of the pendulum can be solved for in terms
of elliptic functions which are generalizations of the trigonometric functions.
4.30. Show that period given in Problem 4.29 can be written as
The binomial theorem states that if \x\ < 1, then
<!+*).> = i + p. + &=$* + "Vffr"  + ■•■
If p = ■—■!■, this can be written
d + aOi/2 = i \ x + ^f*»  f^frf* 3 + •••
Letting x = —k 2 sin 2 <£ and integrating from to v/2, we find
/*ir/2 djp
P  AyjUg I r ,o • o^
J yl — k 2 sin 2 tf>
X ir/2 f 1 1 • 3 1
j 1 +  A; 2 sin 2 ^ + 2T4 ** sin4 * + ' " '  d(f>
= 2.vw{i + (i)'*. + (£!)'*• + (Hfcf)'"' + ••■}
where we have made use of the integration formula
r /2 • a, a l*3'5(2wl) ff
I sin 2 "0^ = 2 .4.6(2n) 2
The term by term integration is possible since fc < 1.
4.31. A bead of mass m is constrained to move on a frictionless wire in the shape of a
cycloid [Fig. 419 below] whose parametric equations are
x = a(cj>  sin </>), y = a(l — cos </>) CO
which lies in a vertical plane. If the bead starts from rest at point O, (a) find the
speed at the bottom of the path and (b) show that the bead performs oscillations
with period equivalent to that of a simple pendulum of length 4a.
CHAP. 4] THE SIMPLE HARMONIC OSCILLATOR AND THE SIMPLE PENDULUM
107
(a)
4.32.
Let P be the position of the bead at any time t
and let s be the arclength along the cycloid meas
ured from point O.
By the conservation of energy, measuring
potential energy relative to line AB through the
minimum point of the cycloid, we have
P.E. at P + K.E. at P = P.E. at O + K.E. at
mg(2ay) + im(ds/dt) 2 = mg(2a) + (2)
Thus
(ds/dt) 2 = 2gy or
ds/dt =
At the lowest point y — 2a the speed is v = \j2g(2a) = 2y/ga.
(b) From part (a), (ds/dt) 2 = 2gy. But
{ds/dt) 2 = (dx/dt) 2 + (dy/dt) 2 = a 2 (l  cos <f>) 2 $ 2 + a 2 sin 2 <f> $ 2 = 2a 2 (l  cos <f>)]> 2
Then 2a 2 (l  cos <f>)^> 2 = 2ga{l — cos </>) or $ 2 = g/a. Thus
dtp/dt = yfgja, and <f> = y/g/at + c x (4)
When <£ = 0, t — 0; when .£ = 2w, t = P/2 where P is the period. Hence from the second
equation of (t),
P = Airya/g — 2wy±a/g
and the period is the same as that of a simple pendulum of length I = 4a.
For some interesting applications see Problems 4.864.88.
A particle of mass m is placed on the inside
of a smooth paraboloid of revolution having
equation cz = x 2 + y 2 at a point P which is at
height H above the horizontal [assumed as the
xy plane]. Assuming that the particle starts
from rest, (a) find the speed with which it
reaches the vertex 0, (b) find the time r taken,
and (c) find the period for small vibrations.
It is convenient to choose the point P in the yz
plane so that x = and cz = y 2 . By the principle
of conservation of energy we have if Q is any point
on the path PQO,
Fig. 420
P.E. at P + K.E. at P = P.E. at Q + K.E. at Q
mgH + £m(0) 2 = mgz + $m(ds/dt) 2
where 8 is the arclength along OPQ measured from O. Thus
(ds/dt) 2 = 2g(Hz)
or ds/dt = — s/2g(H — z)
using the negative sign since s is decreasing with t.
(a) Putting z = 0, we see that the speed is y]2gH at the vertex.
(1)
(6) We have, since x = and cz = y 2 ,
fds\ 2 fdx\* fdyY , fdz\ 2 _ fdyY.W/dy
\dtj \dtj \dtj "*" \dt) ~ \dtj "*" c 2 \dt
Thus (I) can be written (1 + 4y 2 /c 2 )(dy/dt) 2 = 2g(H  y 2 /c). Then
^7 =  y/2gc = or
Vc 2 + 4y2
= <*+Wt)
dt
V"c2 + 4i/2
V2^c dt = , dy
yfcH  y*
108 THE SIMPLE HARMONIC OSCILLATOR AND THE SIMPLE PENDULUM [CHAP. 4
Integrating, using the fact that z = H and thus y = V^H at t = while at t = t, y  0,
we have
f T , f° Vc 2 + 4y2 1 r \^H Vc 2 + % 2
I —y2gc dt = I ■ dy or r = — F == I / <fy
•A) J^ ^/ c # ~ 2/ 2 v20C J VcH  2/2
Letting y = yJcH cos (9, the integral can be written
T = ^— I Vc 2 + 4c# cos 2 9 de = — I Vc 2 + 4cH  AcH sin 2 d*
and this can be written , v/2
r = j£±i«J y/T^W^Jde (3)
where fc = y/AH/{c + AH) < 1 (4)
The integral in (5) is an elliptic integral and cannot be evaluated in terms of elementary
functions. It can, however, be evaluated in terms of series [see Problem 4.119].
(c) The particle oscillates back and forth on the inside of the paraboloid with period given by
p = 4t = 4 J c + m r /2 VI  fe 2 sin 2 » cZg (5)
For small vibrations the value of & given by &) can be assumed so small so as to be zero
for practical purposes. Hence (5) becomes
P = 2w^fJc + A~H)j29
The length of the equivalent simple pendulum is I  ^(c + AH).
Supplementary Problems
SIMPLE HARMONIC MOTION AND THE SIMPLE HARMONIC OSCILLATOR
4.33. A particle of mass 12 gm moves along the x axis attracted toward the point O on it by a force
in dynes which is numerically equal to 60 times its instantaneous distance x cm from O. If the
particle starts from rest at x = 10, find the (a) amplitude, (6) period and (c) frequency of
the motion. Ans. (a) 10 cm, (6) 2jt/Vb sec, (c) Vb/2jt vib/sec
4.34. (a) If the particle of Problem 4.33 starts at x = 10 with a speed toward O of 20 cm/sec, determine
its amplitude, period and frequency. (6) Determine when the particle reaches O for the first time.
Ans. (a) Amplitude = 6^ cm, period = 2^/V5 sec, frequency = V5/2* vib/sec; (b) 0.33 sec
4 35 A particle moves on the x axis attracted toward the origin O on it with a force proportional
to its instantaneous distance from O. If it starts from rest at x = 5 cm and reaches ; x  2.5 cm
for the first time after 2 sec, find (a) the position at any time t after it starts, (6) the speed
at x = 0, (c) the amplitude, period and frequency of the vibration, (d) the maximum acceleration,
(e) the maximum speed.
Ans. (a) a = 5cosM/6); (6) 5W6 cm/sec; (c) 5 cm, 12 sec, 1/12 vib/sec; (d) 5^/36 cm/sec 2 ;
(e) 5?r/6 cm/sec
4 36 If a particle moves with simple harmonic motion along the x axis, prove that (a) the acceleration
is numerically greatest at the ends of the path, (6) the velocity is numerically greatest in the
middle of the path, (c) the acceleration is zero in the middle of the path, (d) the velocity is zero
at the ends of the path.
4 37 A particle moves with simple harmonic motion in a straight line. Its maximum speed is 20 ft/sec
' ' and its maximum acceleration is 80 ft/sec 2 . Find the period and frequency of the motion.
Ans. v/2 sec, 2/v vib/sec
CHAP. 4] THE SIMPLE HARMONIC OSCILLATOR AND THE SIMPLE PENDULUM 109
4.38. A particle moves with simple harmonic motion. If its acceleration at distance D from the
equilibrium position is A, prove that the period of the motion is 2tt^D/A.
4.39. A particle moving with simple harmonic motion has speeds of 3 cm/sec and 4 cm/sec at distances
8 cm and 6 cm respectively from the equilibrium position. Find the period of the motion.
Ans. Av sec
4.40. An 8 kg weight placed on a vertical spring stretches it 20 cm. The weight is then pulled down
a distance of 40 cm and released, (a) Find the amplitude, period and frequency of the oscillations.
(b) What is the position and speed at any time?
Ans. (a) 40 cm, 2ir/l sec, 7/2tt vib/sec
(b) x = 40 cos It cm, v = —280 sin It cm/sec
4.41. A mass of 200 gm placed at the lower end of a vertical spring stretches it 20 cm. When it is in
equilibrium, the mass is hit and due to this goes up a distance of 8 cm before coming down again.
Find (a) the magnitude of the velocity imparted to the mass when it is hit and (6) the period of
the motion. Ans. (a) 56 cm/sec, (6) 2jt/7 sec
4.42. A 5 kg mass at the end of a spring moves with simple harmonic motion along a horizontal straight
line with period 3 sec and amplitude 2 meters, (a) Determine the spring constant. (6) What is the
maximum force exerted on the spring?
Ans. (a) 1140 dynes/cm or 1.14 newtons/meter
(6) 2.28 X 10 5 dynes or 2.28 newtons
4.43. When a mass M hanging from the lower end of a vertical spring i s set into motion, it oscillates
with period P. Prove that the period when mass m is added is Py/l + m/M.
THE DAMPED HARMONIC OSCILLATOR
4.44. (a) Solve the equation d 2 x/dt 2 + 2 dx/dt + 5x  subject to the conditions x = 5, dx/dt = 3
at t — and (6) give a physical interpretation of the results.
Ans. (a) x — J. e t(10 cos 2* 5 sin 2i)
4.45. Verify that the damping force given by equation (2) of Problem 4.11 is correct regardless of the
position and velocity of the particle.
4.46. A 60 lb weight hung on a vertical spring stretches it 2 ft. The weight is then pulled down 3 ft
and released, (a) Find the position of the weight at any time if a damping force numerically
equal to 15 times the instantaneous speed is acting. (6) Is the motion oscillatory damped, over
damped or critically damped? Ans. (a) x  3e«(4£+l), (6) critically damped
4.47. Work Problem 4.46 if the damping force is numerically 18.75 times the instantaneous speed.
Ans. (a) x = 4e~ 2t — e~ 8t , (b) overdamped
4.48. In Problem 4.46, suppose that the damping force is numerically 7.5 times the instantaneous speed.
(a) Prove that the motion is damped oscillatory. (6) Find the amplitude, period and frequency of
the oscillations, (c) Find the logarithmic decrement.
Ans. (b) Amplitude = 2^3 e~^ ft, period = ir/yfz sec, frequency = yfzh vib/sec; (c) 2^/V3
4.49. Prove that the logarithmic decrement is the time required for the maximum amplitude during
an oscillation to reduce to 1/e of this value.
4.50. The natural frequency of a mass vibrating on a spring is 20 vib/sec, while its frequency with
damping is 16 vib/sec. Find the logarithmic decrement. Ans. 3/4
4.51. Prove that the difference in times corresponding to the successive maximum displacements of a
dampe d harmon ic oscillator with equation given by (12) of page 88 is constant and equal to
4nm/\/4/cm — /3 2 .
4.52. Is the difference in times between successive minimum displacements of a damped harmonic
oscillator the same as in Problem 4.51? Justify your answer.
110 THE SIMPLE HARMONIC OSCILLATOR AND THE SIMPLE PENDULUM [CHAP. 4
FORCED VIBRATIONS AND RESONANCE
4.53. The position of a particle moving along the x axis is determined by the equation d 2 x/dt 2 + Adxldt +
8* = 20 cos 2t. If the particle starts from rest at x = 0, find (a) x as a function of t, (b) the
amplitude, period and frequency of the oscillation after a long time has elapsed.
Ans. (a) x = cos 2t + 2 sin 2t  e~ 2t (cos 2t + 3 sin 2t)
(b) Amplitude = yfh, period = v, frequency = 1/v
4.54. (a) Give a physical interpretation to Problem 4.53 involving a mass at the end of a vertical spring.
(b) What is the natural frequency of such a vibrating spring? (c) What is the frequency of the
impressed force? Ans. (b) y/2/ir, (c) 1/tt
4.55. The weight on a vertical spring undergoes forced vibrations according to the equation
d 2 x/dt 2 + 4x = 8 sin at where x is the displacement from the equilibrium position and w > is a
constant. If at t = 0, x = and dx/dt = 0, find (a) x as a function of t, (b) the period of
the external force for which resonance occurs.
Ans. (a) x = (8 sin ut — 4w sin 2t)/(4 — <o 2 ) if w ¥^ 2; x = sin 2t — 2t cos 2t if w = 2
(6) w = 2 or period = it
4.56. A vertical spring having constant 17 lb wt per ft has a 32 1b weight suspended from it. Ah
external force given as a function of time t by F(t) = 65 sin4t, i§0 is applied. A damping
force given numerically in lb wt by 2v, where v is the instantaneous speed of the weight in ft/sec,
is assumed to act. Initially the weight is at rest at the equilibrium position, (a) Determine the
position of the weight at any time. (6) Indicate the transient and steadystate solutions, giving
physical interpretations of each, (c) Find the amplitude, period and frequency of the steadystate
solution. [Use g = 32 ft/sec 2 .]
Ans. (a) x = 4e t cos4t + sin 4t — 4 cos 4*
(6) Transient, 4e t cos4t; steadystate, sin4i — 4 cos4t
(c) Amplitude = y/Vf ft, period = v/2 sec, frequency = 2/tr vib/sec
4.57. A spring is stretched 5 cm by a force of 50 dynes. A mass of 10 gm is placed on the lower end
of the spring. After equilibrium has been reached, the upper end of the spring is moved
up and down so that the external force acting on the mass is given by F(t) = 20 cos at, t§0.
(a) Find the position of the mass at any time, measured from its equilibrium position. (6) Find the
value of (o for which resonance occurs.
Ans. (a) x = (20 cos ut)/(l  u 2 )  20 cos t, (b) u = 1
4.58. A periodic external force acts on a 6 kg mass suspended from the lower end of a vertical spring
having constant 150 newtons/meter. The damping force is proportional to the instantaneous speed
of the mass and is 80 newtons when the speed is 2 meters/sec. Find the frequency at which
resonance occurs. Ans. 5/6v vib/sec
THE SIMPLE PENDULUM
4.59. Find the length of a simple pendulum whose period is 1 second. Such a pendulum which registers
seconds is called a seconds pendulum. Ans. 99.3 cm or 3.26 ft
4.60. Will a pendulum which registers seconds at one location lose or gain time when it is moved to
another location where the acceleration due to gravity is greater? Explain.
Ans. Gain time
4.61. A simple pendulum whose length is 2 meters has its bob drawn to one side until the string makes
an angle of 30° with the vertical. The bob is then released, (a) What is the speed of the bob as
it passes through its lowest point? (6) What is the angular speed at the lowest point? (c) What
is the maximum acceleration and where does it occur?
Ans. (a) 2.93 m/sec, (6) 1.46 rad/sec, (c) 2 m/sec 2
CHAP. 4] THE SIMPLE HARMONIC OSCILLATOR AND THE SIMPLE PENDULUM HI
4.62. Prove that the tension in the string of a vertical simple pendulum of length I and mass m is given
by mg cos e where a is the instantaneous angle made by the string with the vertical.
4.63. A seconds pendulum which gives correct time at a certain location is taken to another location
where it is found to lose T seconds per day. Determine the gravitational acceleration at the second
location. Ans. g(l — T/86,400) 2 where g is the gravitational acceleration at the first location
4.64. What is the length of a seconds pendulum on the surface of the moon where the acceleration due
to gravity is approximately 1/6 that on the earth? Ans. 16.5 cm
4.65. A simple pendulum of length I and mass m hangs vertically from a fixed point O. The bob is given
an initial horizontal velocity of magnitude v . Prove that the arc through which the bob swings
in one period has a length given by 41 cos 1 (1 — v 2 /2gl)
4.66. Find the minimum value of v in Problem 4.65 in order that the bob will make a complete
vertical circle with center at 0. Ans. 2~\fgl
THE TWO AND THREE DIMENSIONAL HARMONIC OSCILLATOR
4.67. A particle of mass 2 moves in the xy plane attracted to the origin with a force given by
F = — 18*i — 50?/j. At t = the particle is placed at the point (3, 4) and given a velocity
of magnitude 10 in a direction perpendicular to the x axis, (a) Find the position and velocity of
the particle at any time. (6) What curve does the particle describe?
Ans. (a) r = 3 cos 3t i + [4 cos ht + 2 sin 5«] j, v = — 9 sin St i + [10 cos 5t  20 sin 5t] j
4.68. Find the total energy of the particle of Problem 4.67. Ans. 581
4.69. A two dimensional harmonic oscillator of mass 2 has potential energy given by V = S(x 2 + 4y 2 ).
If the position vector and velocity of the oscillator at time t = are given respectively by
r = 2i — j and v = 4i + 8j, (a) find its position and velocity at any time t > and (6) deter
mine the period of the motion.
Ans. (a) r = (2 cos At + sin 4t)i + (sin St  cos 8«)j, v = (4 cos 4t  8 sin 4*)i + (8 cos 8i + 8 sin 8*)j
(6) s/8
4.70. Work Problem 4.69 if V = 8(x 2 + 2y 2 ). Is there a period defined for the motion in this case?
Explain.
4.71. A particle of mass m moves in a 3 dimensional force field whose potential is given by
V = ^k(x 2 + 4y 2 + 16z 2 ). (a) Prove that if the particle is placed at an arbitrary point in space
other than the origin, then it will return to the point after some period of time. Determine this
time. (6) Is the velocity on returning to the starting point the same as the initial velocity? Explain.
4.72. Suppose that in Problem 4.71 the potential is V = %k(x 2 + 2y 2 + 5« 2 ). Will the particle return
to the starting point? Explain.
MISCELLANEOUS PROBLEMS
4.73. A vertical spring of constant k having natural length I is supported at a fixed point A. A mass m
is placed at the lower end of the spring, lifted to a height h below A and dropped. Prove that
the lowest point reached will be at a distance below A given by I + mg/ K + y/m 2 g 2 / K 2 + 2mgh/n.
4.74. Work Problem 4.73 if damping proportional to the instantaneous velocity is taken into account.
4.75 Given the equation mx + /3x + kx = for damped oscillations of a harmonic oscillator. Prove that
if E = \mx 2 + \kx 2 , then E = p'x 2 . Thus show that if there is damping the total energy E
decreases with time. What happens to the energy lost? Explain.
112 THE SIMPLE HARMONIC OSCILLATOR AND THE SIMPLE PENDULUM [CHAP. 4
4.76. (a) Prove that A x cos (at  4> x ) + A 2 cos (at  <p 2 ) = A cos (at  </>)
/ Al sin^t + A 2 sin^> 2 \
where A = y/A\ + A\ + 2A 1 A 2 cos (^  <p 2 ), <fi = ^'(^cos^ + A 2 cos <t>J '
(6) Use (a) to demonstrate that the sum of two simple harmonic motions of the same frequency
and in the same straight line is simple harmonic of the same frequency.
4.77. Give a vector interpretation to the results of Problem 4.76.
4.78. Discuss Problem 4.76 in case the frequencies of the two simple harmonic motions are not equal.
Is the resultant motion simple harmonic? Justify your answer.
4.79. A particle oscillates in a plane so that its distances x and y from two mutually perpendicular
axes are given as functions of time t by
x = A cos (at + X ), y = B cos (at + </> 2 )
(a) Prove that the particle moves in an ellipse inscribed in the rectangle defined by x = ±A,
y — ±b (&) p r ove that the period of the particle in the elliptical path is 2ir/a.
4.80. Suppose that the particle of Problem 4.79 moves so that
X = A COS (at + 0i), y — B COS (at + et + 2 )
where « is assumed to be a positive constant which is assumed to be much smallertiian a. Vrvve
that the particle oscillates in slowly rotating ellipses inscribed in the rectangle x  ±A, y  ±B.
4.81. Illustrate Problem 4.80 by graphing the motion of a particle which moves in the path
x = 3 cos(2t + n/4), y = 4 cos (2.4t)
nmr
m
K 2
^TOKT^
4.82. In Fig. 421 a mass m which is on a frictionless
table is connected to fixed points A and B by
two springs of equal natural length, of negli
gible mass and spring constants /q and /c 2 re ^
spectively. The mass m is displaced horizontally A B
and then released. Prove that the perio d of
oscillation is given by P = 2v\/'m/(K 1 + k 2 ). Fig. 421
4.83. A spring having constant k and negligible mass has
one end fixed at point A on an inclined plane of
angle a and a mass m at the other end, as indicated
in Fig. 422. If the mass m is pulled down a distance
x below the equilibrium position and released, find
the displacement from the equilibrium position at any
time if (a) the incline is frictionless, (6) the incline
has coefficient of friction /i. Fig. 42
4.84. A particle moves with simple harmonic motion along the x axis. At times t , 2t Q an dJ*o i* ™
located at * ■ = a, b and c respectively. Prove that the period of oscillation is cos i (a + c )/26*
4 85. A seconds pendulum giving the correct time at one location is taken to another locati or i where
it loses 5 minutes per day. By how much must the pendulum rod be lengthened or shortened in
order to give the correct time?
4.86. A vertical pendulum having a bob of mass m is sus
pended from the fixed point O. As it oscillates, the
string winds up on the constraint curves OB A [or OC\
as indicated in Fig. 423. Prove that if curve ABC is a
cycloid, then the period of oscillation will be the same
regardless of the amplitude of the oscillations. The pen
dulum in this case is called a cycloidal pendulum. The
curves ODA and OC are constructed to be evolutes of
the cycloid. [Hint. Use Problem 4.31.] Iig.4A»
CHAP. 4] THE SIMPLE HARMONIC OSCILLATOR AND THE SIMPLE PENDULUM 113
4.87. A bead slides down a frictionless wire located in a vertical plane. It is desired to find the
shape of the wire so that regardless of where the bead is placed on the wire it will slide under
the influence of gravity to the bottom of the wire in the same time. This is often called the
tautochrone problem. Prove that the wire must have the shape of a cycloid.
[Hint. Use Problem 4.31.]
4.88. Prove that the curves ODA and OC of Problem 4.86 are cycloids having the same shape as the
cycloid ABC.
4.89. A simple pendulum of length I has its point of support moving back and forth on a horizontal line
so that its distance from a fixed point on the line is A sin ut, t ^ 0. Find the position of the
pendulum bob at any time t assuming that it is at rest at the equilibrium position at * = 0.
4.90. Work Problem 4.89 if the point of support moves vertically instead of horizontally and if at
t — the rod of the pendulum makes an angle O with the vertical.
4.91. A particle of mass m moves in a plane under the influence of forces of attraction toward fixed
points which are directly proportional to its instantaneous distance from these points. Prove
that in general the particle will describe an ellipse.
4.92. A vertical elastic spring of negligible weight and having its upper end fixed, carries a weight
W at its lower end. The weight is lifted so that the tension in the spring is zero, and then it is
released. Prove that the tension in the spring will not exceed 2W.
4.93. A vertical spring having constant /c has a pan on top of it with
a weight W on it [see Fig. 424]. Determine the largest fre W
quency with which the spring can vibrate so that the weight
will remain in the pan. ,
4.94. A spring has a natural length of 50 cm and a force of 100 dynes
is required to stretch it 25 cm. Find the work done in stretching
the spring from 75 cm to 100 cm, assuming that the elastic limit
is not exceeded so that the spring characteristics do not change.
Ans. 3750 ergs
4.95. A particle moves in the xy plane so that its position is given by
x = Acosw«, y = B cos 2ut. Prove that it describes an arc of a
parabola. Fig. 4.24
4.96. A particle moves in the xy plane so that its position is given by x = A cos (u^ + 9^),
y = Bcos(w 2 t + <f> 2 ) Prove that the particle describes a closed curve or not, according as a t /u 2 is
rational or not. In which cases is the motion periodic?
4.97. The position of a particle moving in the xy plane is described by the equations d 2 x/dt 2 = —Ay,
d 2 y/dt 2 = 4x. At time t = the particle is at rest at the point (6, 3). Find (a) its position
and (b) its velocity at any later time t.
4.98. Find the period of a simple pendulum of length 1 meter if the maximum angle which the rod
makes with the vertical is (a) 30°, (6) 60°, (c) 90°.
4.99. A simple pendulum of length 3 ft is suspended vertically from a fixed point. At t = the bob is
given a horizontal velocity of 8 ft/sec. Find (a) the maximum angle which the pendulum rod
makes with the vertical, (b) the period of the oscillations.
Ans. (a) cos" 1 2/3 = 41° 48', (6) 1.92 sec
4.100. Prove that the time averages over a period of the potential energy and kinetic energy of a
simple harmonic oscillator are equal to 2tt 2 A 2 IP 2 where A is the amplitude and P is the period
of the motion.
4.101. A cylinder of radius 10 ft with its axis vertical oscillates vertically in water of density 62.5 lb/ft 3
with a period of 5 seconds. How much does it weigh? Ans. 3.98Xl0 5 lbwt
4.102. A particle moves in the xy plane in a force field whose potential is given by V = x 2 + xy + y 2 .
If the particle is initially at the point (3, 4) and is given a velocity of magnitude 10 in a direction
parallel to the positive x axis, (a) find the position at any time and (6) determine the period of
the motion if one exists.
114 THE SIMPLE HARMONIC OSCILLATOR AND THE SIMPLE PENDULUM [CHAP. 4
4.103. In Problem 4.96 suppose that a x /a 2 is irrational and that at t — the particle is at the
particular point (x ,y ) inside the rectangle denned by x = ±A, y — ±B. Prove that the point
(x ,y ) will never be reached again but that in the course of its motion the particle will come
arbitrarily close to the point.
4.104. A particle oscillates on a vertical frictionless cycloid with its vertex downward. Prove that the
projection of the particle on a vertical axis oscillates with simple harmonic motion.
4.105. A mass of 5 kg at the lower end of a vertical spring which has an elastic constant equal to
20 newtons/meter oscillates with a period of 10 seconds. Find (a) the damping constant, (6) the
natural period and (c) the logarithmic decrement. Ans. (a) 19 nt sec/m, (6) 3.14 sec
4.106. A mass of 100 gm is supported in equilibrium by two identical
springs of negligible mass having elastic constant equal to
50 dynes/cm. In the equilibrium position shown in Fig. 425
the springs make an angle of 30° with the horizontal and are
100 cm in length. If the mass is pulled down a distance of
2 cm and released, find the period of the resulting oscillation.
4.107. A thin hollow circular cylinder of inner radius 10 cm is fixed
so that its axis is horizontal. A particle is placed on the inner
frictionless surface of the cylinder so that its vertical distance
above the lowest point of the inner surface is 2 cm. Find
(a) the time for the particle to reach the lowest point and
(b) the period of the oscillations which take place.
4.108. A cubical box of side a and weight W vibrates vertically in water of density o. Prove that the
period of vibration is {2ir/a)^<jg/W.
4.109. A spring vibrates so that its equation of motion is
md 2 x/dt* + kx = F(t)
If x = 0, dx/dt  at t = 0, find it as a function of time t.
i r* i —
Ans. x = — — I F(u) sin V «/m (t — u) du
■yrriK ^o
4.110. Work Problem 4.109 if damping proportional to dx/dt is taken into account.
4.111. A spring vibrates so that its equation of motion is
m d 2 x/dt 2 + kx — 5 cos cot + 2 cos 3wt
If x = 0, x = v at t = 0, (a) find x at any time * and (6) determine for what values of a
resonance will occur.
4 112 A vertical spring having elastic constant k carries a mass m at its lower end. At t = the
spring is in equilibrium and its upper end is suddenly made to move vertically so that its distance
from the original point of support is given by A sin a t, t § 0. Find (a) the position of the mass w
at any time and (b) the values of <o for which resonance occurs.
4.113. (a) Solve d*x/dt* + x = t sin t + cos t where x = 0, dx/dt = at t = 0, and (6) give a physical
interpretation.
4.114. Discuss the motion of a simple pendulum for the case where damping and external forces are
present.
CHAP. 4] THE SIMPLE HARMONIC OSCILLATOR AND THE SIMPLE PENDULUM 115
4.115. Find the period of small vertical oscillations of a cylinder of radius a and height h floating
with its axis horizontal in water of density a.
4.116. A vertical spring having elastic constant 2 newtons per meter has a 50 gm weight suspended
from it. A force in newtons which is given as a function of time t by F(t) = 6 cos 4 t, t ^
is applied. Assuming that the weight, initially at the equilibrium position, is given an upward
velocity of 4 m/sec and that damping is negligible, determine the (a) position and (6) velocity
of the weight at any time.
4.117. In Problem 4.55, can the answer for « = 2 be deduced from the answer for « ^ 2 by taking
the limit as u »• 2? Justify your answer.
4.118. An oscillator has a restoring force acting on it whose magnitude is — kx — ex 2 where e is small
compared with k. Prove that the displacement of the oscillator [in this case often called an
anharmonic oscillator] from the equilibrium position is given approximately by
A 2
x = A cos (ut<f>) + ^—{cos2(ut — d>)3}
OK
where A and </> are determined from the initial conditions.
4.119. Prove that if the oscillations in Problem 4.32 are not necessarily small, then the period is given by
Chapter 5
CENTRAL FORCES
and PLANETARY MOTION
CENTRAL FORCES
Suppose that a force acting on a particle of
mass m is such that [see Fig. 51]:
(a) it is always directed from m toward or
away from a fixed point O,
(b) its magnitude depends only on the distance
r from O.
Then we call the force a central force or central
force field with as the center of force. In sym
bols F is a central force if and only if
F = f(r)n = f(r)r/r (1) Fig. 51
where ri = r/r is a unit vector in the direction of r.
The central force is one of attraction toward O or repulsion from O according as
f(r) < or f(r) > respectively.
SOME IMPORTANT PROPERTIES OF CENTRAL FORCE FIELDS
If a particle moves in a central force field, then the following properties are valid.
1. The path or orbit of the particle must be a plane curve, i.e. the particle moves in
a plane. This plane is often taken to be the xy plane. See Problem 5.1.
2. The angular momentum of the particle is conserved, i.e. is constant. See Problem 5.2.
3. The particle moves in such a way that the position vector or radius vector drawn
from to the particle sweeps out equal areas in equal times. In other words, the
time rate of change in area is constant. This is sometimes called the law of areas.
See Problem 5.6.
EQUATIONS OF MOTION FOR A PARTICLE
IN A CENTRAL FIELD
By Property 1, the motion of a particle in a cen
tral force field takes place in a plane. Choosing this
plane as the xy plane and the coordinates of the par
ticle as polar coordinates (r, 6), the equations of mo
tion are found to be [see Problem 5.3]
m(rr¥) = f{r) (2)
m(r'e+ 2r9) = (8)
where dots denote differentiations with respect to
time t.
(r,»)
Fig. 52
116
CHAP. 5] CENTRAL FORCES AND PLANETARY MOTION 117
From equation (3) we find
r 2 6 = constant = h (4)
This is related to Properties 2 and 3 above.
IMPORTANT EQUATIONS DEDUCED FROM
THE EQUATIONS OF MOTION
The following equations deduced from the fundamental equations (2) and (3) often
prove to be useful.
1 •• h 2 f(r)
!■• T — — = ■ (K\
r 3 m * '
d 2 u 1
where u = 1/r.
d0 2 r\de) r mh 2 ^
POTENTIAL ENERGY OF A PARTICLE IN A CENTRAL FIELD
A central force field is a conservative field, hence it can be derived from a potential.
This potential which depends only on r is, apart from an arbitrary additive constant,
given by
V(r) = jf(r)dr (g)
This is also the potential energy of a particle in the central force field. The arbitrary
additive constant can be obtained by assuming, for example, 7 = at r = or V *
as r* oo.
CONSERVATION OF ENERGY
By using (8) and the fact that in polar coordinates the kinetic energy of a particle is
im^ + r^e 2 ), the equation for conservation of energy can be written
im(r 2 + r 2 6 2 ) + V(r) = E (P)
or \m(r 2 + r 2 6 2 )  f f(r) dr = E (io)
where E is the total energy and is constant. Using (4), equation (10) can also be written as
mh 2 Vfdr\ 2 „1 f „ , ,
and also as f ( f2 + S) ~ J ^ dr = E W
In terms of u = 1/r, we can also write equation (9) as
f^V ^ . 2 _ 2(E~V)
KdeJ + u = mh 2 (I 3 )
DETERMINATION OF THE ORBIT FROM THE CENTRAL FORCE
;d, i.e. if f(r) is given, it is possi
rbit can be obtained in the form
r = r{0) ( U )
If the central force field is prescribed, i.e. if f(r) is given, it is possible to determine
the orbit or path of the particle. This orbit can be obtained in the form
118
CENTRAL FORCES AND PLANETARY MOTION
[CHAP. 5
i.e. r as a function of 9, or in the form
r = r{t), 9 = 9{t) (IS)
which are parametric equations in terms of the time parameter t.
To determine the orbit in the form (U) it is convenient to employ equations (6), (7)
or (11). To obtain equations in the form (15), it is sometimes convenient to use (12) together
with (A) or to use equations (4) and (5).
DETERMINATION OF THE CENTRAL FORCE FROM THE ORBIT
Conversely if we know the orbit or path of the particle, then we can find the correspond
ing central force. If the orbit is given by r = r(9) or u = u(9) where u = 1/r, the central
force can be found from
mh 2 [d 2 r 2 /dr\ 2 _
,de)
f(r) =
or
f(Vu) =
de*
mh 2 u 2
(16)
d 2 u
de 2
+ u
(17)
which are obtained from equations (6) and (7) on page 117. The law of force can also be
obtained from other equations, as for example equations (9)(13).
It is important to note that given an orbit there may be infinitely many force fields for
which the orbit is possible. However, if a central force field exists it is unique, i.e. it is
the only one.
CONIC SECTIONS, ELLIPSE, PARABOLA AND HYPERBOLA
Consider a fixed point and a fixed line AB distant D from 0, as shown in Fig. 53.
Suppose that a point P in the plane of and AB moves so that the ratio of its distance
from point to its distance from line AB is always equal to the positive constant € .
Then the curve described by P is given in
polar coordinates (r, 9) by
V
r —
(18)
1 + e COS 9
See Problem 5.16.
The point is called a focus, the line AB is
called a directrix and the ratio e is called the
eccentricity. The curve is often called a conic
section since it can be obtained by intersecting
a plane and a cone at different angles. Three
possible types of curves exist, depending on the
value of the eccentricity.
1
1
A
^
^v
X
V
^ Dir
p
\p d
M
r/\
Focus \^
A' \
/
/
/
/
E
/
/
m u »
S
^
B
Fig. 53
Ellipse: « < 1 [See Fig. 54 below.]
If C is the center of the ellipse and CV = CU = a is the length of the semimajor
axis, then the equation of the ellipse can be written as
a(le 2 )
r i + € cos e
(19)
Note that the major axis is the line joining the vertices V and U of the ellipse and has
length 2a.
CHAP. 5]
CENTRAL FORCES AND PLANETARY MOTION
119
If b is the length of the semiminor axis
[CW or CS in Fig. 54] and c is the distance
CO from center to focus, then we have the
important result
c = \Za 2  b 2 = a c (20)
A circle can be considered as a special case
of an ellipse with eccentricity equal to zero.
2. Parabola: £ = 1 [See Fig. 55.]
The equation of the parabola is
r =
P
1 + cos
(21)
We can consider a parabola to be a
limiting case of the ellipse (19) where « * 1,
which means that a » °o [i.e. the major
axis becomes infinite] in such a way that
a(l  £ 2 ) = p.
Hyperbola: e > 1 [See Fig. 56.]
The hyperbola consists of two branches
as indicated in Fig. 56. The branch on the
left is the important one for our purposes.
The hyperbola is asymptotic to the dashed
lines of Fig. 56 which are called its asymp
totes. The intersection C of the asymptotes
is called the center. The distance CV = a
from the center C to vertex V is called the
semimajor axis [the major axis being the
distance between vertices V and U by anal
ogy with the ellipse]. The equation of the
hyperbola can be written as
r = a(< 2  l)
1 + e cos 9
(22)
Fig. 55
Fig. 56
Various other alternative definitions for conic sections may be given. For example, an
ellipse can be defined as the locus or path of all points the sum of whose distances from two
fixed points is a constant. Similarly, a hyperbola can be defined as the locus of all points
the difference of whose distances from two fixed points is a constant. In both these cases
the two fixed points are the foci and the constant is equal in magnitude to the length of
the major axis.
SOME DEFINITIONS IN ASTRONOMY
A solar system is composed of a star [such as our sun] and objects called planets which
revolve around it. The star is an object which emits its own light, while the planets do
not emit light but can reflect it. In addition there may be objects revolving about the
planets. These are called satellites.
In our solar system, for example, the moon is a satellite of the earth which in turn is a
planet revolving about our sun. In addition there are artificial or manmade satellites
which can revolve about the planets or their moons.
120 CENTRAL FORCES AND PLANETARY MOTION [CHAP. 5
The path of a planet or satellite is called its orbit. The largest and smallest distances
of a planet from the sun about which it revolves are called the aphelion and perihelion
respectively. The largest and smallest distances of a satellite around a planet about which
it revolves are called the apogee and perigee respectively.
The time for one complete revolution of a body in an orbit is called its period. This is
sometimes called a sidereal period to distinguish it from other periods such as the period
of earth's motion about its axis, etc.
KEPLER'S LAWS OF PLANETARY MOTION
Before Newton had enunciated his famous laws /^ ^N^pianet
of motion, Kepler, using voluminous data accumu
lated by Tycho Brahe formulated his three laws
concerning the motion of planets around the sun
[see Fig. 57].
1. Every planet moves in an orbit which is an
ellipse with the sun at one focus. Fig. 57
2. The radius vector drawn from the sun to any planet sweeps out equal areas in
equal times (the law of areas, as on page 116).
3. The squares of the periods of revolution of the planets are proportional to the cubes
of the semimajor axes of their orbits.
NEWTON'S UNIVERSAL LAW OF GRAVITATION
By using Kepler's first law and equations (16) or (17), Newton was able to deduce his
famous law of gravitation between the sun and planets, which he postulated as valid for any
objects in the universe [see Problem 5.21].
Newton's Law of Gravitation. Any two particles of mass mi and m 2 respectively and
distance r apart are attracted toward each other with a force
F = _ Gmim 2 ri ^3)
where G is a universal constant called the gravitational constant.
By using Newton's law of gravitation we can, conversely, deduce Kepler's laws [see
Problems 5.13 and 5.23]. The value of G is shown in the table on page 342.
ATTRACTION OF SPHERES AND OTHER OBJECTS
By using Newton's law of gravitation, the forces of attraction between large objects
such as spheres can be determined. To do this, we use the fact that each large object is
composed of particles. We then apply the law of gravitation to find the forces between
particles and sum over these forces, usually by methods of integration, to find the resultant
force of attraction. An important application of this is given in the following
Theorem 5.1. Two solid or hollow uniform spheres of masses mi and m 2 respectively
which do not intersect are attracted to each other as if they were particles of the same
mass situated at their respective geometric centers.
Since the potential corresponding to
F = ^^ri (24)
CHAP. 5] CENTRAL FORCES AND PLANETARY MOTION 121
is V = Q™»* {25)
r
it is also possible to find the attraction between objects by first finding the potential and
then using F = y7. See Problems 5.265.33.
MOTION IN AN INVERSE SQUARE FORCE FIELD
As we have seen, the planets revolve in elliptical orbits about the sun which is at one
focus of the ellipse. In a similar manner, satellites (natural or manmade) may revolve around
planets in elliptical orbits. However, the motion of an object in an inverse square field of
attraction need not always be elliptical but may be parabolic or hyperbolic. In such cases
the object, such as a comet or meteorite, would enter the solar system and then leave but
never return again.
The following simple condition in terms of the total energy E determines the path of
an object.
(i) if E < the path is an ellipse
(ii) if E = the path is a parabola
(iii) if E > the path is a hyperbola
Other conditions in terms of the speed of the object are also available. See Problem 5.37.
In this chapter we assume the sun to be fixed and the planets do not affect each other.
Similarly in the motion of satellites around a planet such as the earth, for example, we
assume the planet fixed and that the sun and all other planets have no effect.
Although such assumption is correct as a first approximation, the influence of other
planets may have to be taken into account for more accurate purposes. The problems of
dealing with the motions of two, three, etc., objects under their mutual attractions are often
called the two body problem, three body problem, etc.
Solved Problems
CENTRAL FORCES AND IMPORTANT PROPERTIES
5.1. Prove that if a particle moves in a central force field, then its path must be a plane
curve.
Let F = f(r) r x be the central force field. Then
r X F = f(r) r X r x = (1)
since r 1 is a unit vector in the direction of the position vector r. Since F = mdv/dt, this can be
written
r X dx/dt = (2)
or (rXv) = (3)
Integrating, we find r X v = h (4)
where h is a constant vector. Multiplying both sides of (4) by r • ,
r«h = (5)
using the fact that r • (r X v) = (r X r) • v = 0. Thus r is perpendicular to the constant vector h,
and so the motion takes place in a plane. We shall assume that this plane is taken to be the
xy plane whose origin is at the center of force.
122
CENTRAL FORCES AND PLANETARY MOTION
[CHAP. 5
5.2. Prove that for a particle moving in a central force field the angular momentum is
conserved.
From equation (4) of Problem 5.1, we have
r X v = h
where h is a constant vector. Then multiplying by mass m,
w(r X v) = mh (1)
Since the left side of (1) is the angular momentum, it follows that the angular momentum is
conserved, i.e. is always constant in magnitude and direction.
EQUATIONS OF MOTION FOR A PARTICLE IN A CENTRAL FIELD
5.3. Write the equations of motion for a particle in a central field.
By Problem 5.1 the motion of the particle takes place in a plane. Choose this plane to be
the xy plane and the coordinates describing the position of the particle at any time t to be
polar coordinates (r, e). Using Problem 1.49, page 27, we have
(mass) (acceleration) = net force
m{(r — re 2 )r 1 + {r'o + 2r'e)0i} = f( r ) r i (■*)
Thus the required equations of motion are given by
m(r  re 2 ) = f{r) (2)
m(r m e + 2r'e)  (8)
5.4. Show that r 2 6 = h, a constant.
Method 1. Equation (3) of Problem 5.3 can be written
m
m(r e + 2re)
r
(r 2 e + 2rre)
m d_
~r di
(r 2 e)
Thus
dt
(r 2 e) — and so
r 2
(1)
where h is a constant.
Method 2. By Problem 1.49, page 27, the velocity in polar coordinates is
v = frj + rbei
Then from equation (4) of Problem 5.1
h = r X v = f(r X r x ) + ro(r X oj = r 2 tfk (2)
since r X r t = and r X $ x — rk where k is the unit vector in a direction perpendicular to the
plane of motion [the xy plane], i.e. in the direction r X v. Using h = fik in (2), we see that
r 2 e — h.
5.5. Prove that r 2 & = 2A where A is the time rate
at which area is swept out by the position
vector r.
Suppose that in time At the particle moves from
M to N [see Fig. 58]. The area AA swept out by the
position vector in this time is approximately half the
area of a parallelogram with sides r and Ar or [see
Problem 1.18, page 15]
AA = ArXAr
Dividing by Ai and letting
AA
lim
At»0 At
lim 
AttO 2
A«*0,
Ar I
At I
r X
r X v
Fig. 58
CHAP. 5] CENTRAL FORCES AND PLANETARY MOTION 123
i.e., A = irXv = \r 2
using the result in Problem 5.4. Thus r 2 o  2A, as required. The vector quantity
A = Ak = i(rXv) = ^(r 2 *)k
is often called the areal velocity.
5.6. Prove that for a particle moving in a central force field the areal velocity is constant.
By Problem 5.4, r 2 = h = a constant. Then the areal velocity is
A = ±r 2 6k — \hk = ^h, a constant vector
The result is often stated as follows: If a particle moves in a central force field with O as
center, then the radius vector drawn from O to the particle sweeps out equal areas in equal
times. This result is sometimes called the law of areas.
5.7. Show by means of the substitution r = 1/u that the differential equation for the
path of the particle in a central field is
d 2 u , f(l/u)
+ u — 
do 2 mh 2 u 2
From Problem 5.4 or equation (S) of Problem 5.3, we have
r 2 o = h or = h/r 2 = hu 2 (1)
Substituting into equation (2) of Problem 5.3, we find
m(f h 2 /r 3 ) = f(r) (2)
Now if r — 1/u, we have
dr _ dr cfo _ h_ dr_ _ _vdu
dt ~ de dt ~ r 2 do do
r = tt = ^^r = z— = *~ (5)
dr d ( ,du\ d ( , du\ do _ t,2..2 d2u //\
r = Tt = dt\ h Te) = de\r h Te)dt ~ " W ^ (4)
From this we see that (2) can be written
m(h 2 u 2 d 2 u/do 2  h 2 u?) = f(l/u) (5)
or, as required, ^ + u = " ^f W
POTENTIAL ENERGY AND CONSERVATION OF ENERGY
FOR CENTRAL FORCE FIELDS
5.8. (a) Prove that a central force field is conservative and (b) find the corresponding
potential energy of a particle in this field.
Method 1.
If we can find the potential or potential energy, then we will have also incidentally proved
that the field is conservative. Now if the potential V exists, it must be such that
Frfr = dV (1)
where F — f(r) r x is the central force. We have
F • dr = f(r) r t • dr = f(r) • dr = f(r) dr
since r • dr — r dr.
Since we can determine V such that
dV = f(r)dr
124 CENTRAL FORCES AND PLANETARY MOTION [CHAP. 5
for example, V =  J f(r) dr (2)
it follows that the field is conservative and that (2) represents the potential or potential energy.
Method 2.
We can show that V X F = directly, but this method is tedious although straightforward.
5.9. Write the conservation of energy for a particle of mass m in a central force field.
Method 1. The velocity of a particle expressed in polar coordinates is [Problem 1.49, page 27]
v = rr t + reOi s0 tnat y2 = v * v = r2 "*" r2 * 2
Then the principle of conservation of energy can be expressed as
imt) 2 + V = E or im(rHrV)  J f(r)dr  E
where £ is a constant.
Method 2. The equations of motion for a particle in a central field are, by Problem 5.3,
m(r — re 2 ) — f(r) (1)
m(r 6 + 2re) = (2)
Multiply equation (1) by r, equation (2) by re and add to obtain
m(r r + r 2 o + rre 2 ) = f(r)r (3)
This can be written
lm^(r 2 + r 2 'e 2 ) = ~ f f(r) dr U)
Then integrating both sides, we obtain
i m(r 2 + r 2 'e 2 )  J f(r) dr = E (5)
5.10. Show that the differential equation describing the motion of a particle in a central
field can be written as
™ h2 ^\ 2 + 1 *]ff(r)dr = E
2r 4
l\dej
From Problem 5.9 we have by the conservation of energy,
m(r 2 + r 2 e 2 )  J f(r) dr = E (1)
_ dr _ dr do _ dr •
We also have r ~ dt ~ Tedt ~ ~d6°
Substituting {2) into (1), we find
(5) , + ">J'«"* = E ° r l£[(£) 2+r2 ]J' w * = E
since e — h/r 2 .
5.11. (a) If u=l/r, prove that v 2 = f 2 + W 2 = ^ 2 {(d%/^) 2 + u 2 }.
(b) Use (a) to prove that the conservation of energy equation becomes
(du/dOf + v? = 2(EV)/mh 2
CHAP. 5] CENTRAL FORCES AND PLANETARY MOTION 125
(a) From equations (i) and (3) of Problem 5.7 we have 8 = hu 2 , r = —hdu/de. Thus
v 2 = ' r 2 + r 2^2 = hHdu/de) 2 + (l/u 2 )(hu 2 ) 2 = h 2 {(du/de) 2 + u 2 }
(b) From the conservation of energy [Problem 5.9] and part (a),
x mv 2 = i m (f2 + r ' e 2) = e V or (du/de) 2 + u 2 = 2(E  V)/mh 2
DETERMINATION OF ORBIT FROM CENTRAL FORCE,
OR CENTRAL FORCE FROM ORBIT
5.12. Show that the position of the particle as a function of time t can be determined
from the equations
t = J* [G(r)]"*dr, t =  f r 2 d6
where G(r) = — + — f f(r) dr  ^^
v ' m m J JK ' 2m 2 r 2
Placing $ — h/r 2 in the equation for conservation of energy of Problem 5.9,
%m(r 2 + h 2 /r 2 )  f f(r) dr = E
or r 2 = — +  f f(r) dr %  G(r)
m m J JS ' r 2 '
Then assuming the positive square root, we have
dr/dt = y/Gir)
and so separating the variables and integrating, we find
t = f [G(r)]v*dr
The second equation follows by writing e — h/r 2 as dt — r 2 de/h and integrating.
5.13. Show that if the law of central force is denned by
/(r) = Kir 2 , K>0
i.e. an inverse square law of attraction, then the path of the particle is a conic.
Method 1.
In this case f(l/u) = —Ku 2 . Substituting into the differential equation of motion in Problem 5.7,
we find
d 2 u/de 2 + u = K/mh 2 (l)
This equation has the general solution
u = A cos 8 4 B sin + K/mh 2 (2)
or using Problem 4.2, page 92,
u = K/mh 2 + C cos {o — <p) {$)
_ 1
1,e " r K/mh 2 + C cos (e  0) W
It is always possible to choose the axes so that <j> = 0, in which case we have
K/mh 2 + Ccoss
126 CENTRAL FORCES AND PLANETARY MOTION [CHAP. 5
This has the general form of the conic [see Problem 5.16]
P = 1
T 1 + e COS 6 1/p + (e/p) COS
(6)
Then comparing (5) and (6) we see that
1/p = tf/mft 2 , e/p = C (7)
or p = m/i 2 /K, e = mh 2 C/K (8)
Method 2. Since /(r) = K/r 2 , we have
V =  £ f(r) dr = K/r + c x (9)
where c x is a constant. If we assume that V » as r > °°, then Cj = and so
V = ff/r (10)
Using Problem 5.10, page 124, we find
mh 2 /dr\ 2
2r 4
dej +r2
E +  (11)
r
from which To = ±r \^^ ( }
By separating variables and integrating [see Problem 5.66] we find the solution (5) where C is
expressed in terms of the energy E.
5.14. (a) Obtain the constant C of Problem 5.13 in terms of the total energy E and (b) thus
show that the conic is an ellipse, parabola or hyperbola according as E < 0, E = 0,
E > respectively.
Method 1.
(a) The potential energy is
v =  f f(r) dr = f (K/r2) dr = K/r = Ku (1)
where we use u = \/r and choose the constant of integration so that lim V = 0. Now from
equation (5) of Problem 5.13,
u  1/r = K/mh 2 + C cos e (2)
Thus from Problem 5.11(6) together with (1), we have
/ K „ V 2E , 2K ( K . n
(Cs^e^ + i^+C cose) = ^ + ^p(^p + C cos.
K 2 2E \ K 2 2lT
c2 = ^p + rf or c = V^ + ^ w
(6) Using the value of C in part (a), the equation of the conic becomes
or
assuming C > 0.
{■♦V
Comparing this with (4) of Problem 5.16, we see that the eccentricity is
1 K L , , 2ffmft 2
= ^1+^ (4)
#2
From this we see that the conic is an ellipse if £7 < [but greater than K»/2mfc»}, a parabola
if E = and a hyperbola if E > 0, since in such cases e < 1, e1 and e>l respectively.
Method 2. The value of C can also be obtained as in the second method of Problem 5.13.
CHAP. 5]
CENTRAL FORCES AND PLANETARY MOTION
127
5.15. Under the influence of a central force at point 0, a particle moves in a circular orbit
which passes through O. Find the law of force.
Method 1.
In polar coordinates the equation of a circle of radius a
passing through O is [see Fig. 59]
r = 2a cos 9
Then since u = 1/r = (sec 8)/2a, we have
du _ sec e tan e
de ~ 2a
d 2 u _ (sec fl)(sec 2 e) + (sec 6 tan fl)(tan e)
de 2 ~~ 2a
sec 3 9 + sec 9 tan 2 9
2a
Thus by Problem 5.7,
cPu
f(l/u) = mh 2 u 2 ( ^f + u j = mh 2 u 2
Fig. 59
/ sec 3 9 + sec 8 tan 2 e + sec e
mh 2 u 2
2a
8mh 2 a 2 u 5
{sec 3 + sec (tan 2 + 1)} =
,. . 8m h 2 a 2
/(»") = IS—
2a
mh 2 u 2
2a
2 sec 3 e
Thus the force is one of attraction varying inversely as the fifth power of the distance from O.
Method 2. Using r — 2a cos e in equation (16), page 118, we have
4amh 2 SaHnh 2
mh 2 2
— 7 \ —2a cos e — ^ (—2a sin 0) 2 — 2a cos 9
r 4 2a cos 9
r 4 cos 9
CONIC SECTIONS. ELLIPSE, PARABOLA AND HYPERBOLA
5.16. Derive equation (18), page 118, for a conic section.
Referring to Fig. 53, page 118, by definition of a conic section we have for any point P on it,
rid = e or d = r/e (1)
Corresponding to the particular point Q, we have
p/D = e or p = eD (2)
But
t r
D — d + r cos 9 = — V r cos 9 = — (1 + e cos 8)
Then from (2) and (3), we have on eliminating D,
p = r(l + e cos 9) or r
V
1 + e COS
(5)
(4)
The equation is a circle if e = 0, an ellipse if < e < 1, a parabola if e = 1 and a hyperbola
if e > 1.
5.17. Derive equation (19), page 118, for an ellipse.
Referring to Fig. 54, page 119, we see that when 9 = 0, r = OV and when 9 = v,
r = OU. Thus using equation (4) of Problem 5.16,
128 CENTRAL FORCES AND PLANETARY MOTION [CHAP. 5
OV = p/(l + e), OU = p/(le) (0
But since 2a is the length of the major axis,
OV + OU = 2a or p/(l + e) + p/(l  e) = 2a (*)
from which V = a(l  e 2 ) (3)
Thus the equation of the ellipse is
o(l  e 2 )
1 + e COS 9
U)
5.18. Prove that in Fig. 54, page 119, (a) OV = a(l  e ), (b) OU = a(l + e).
(a) From Problem 5.17, equation (5) and the first equation of (1),
OV = ^ = 4^ = «( 1 " e ) (i)
1 + e 1 + e
(b) From Problem 5.17, equation (3) and the second equation of (1),
ou = ^ = ^r^ = a < 1 + e > (a)
1 — e 1—6
5.19. Prove that c = a £ where c is the distance from the center to the focus of the ellipse.
a is the length of the semimajor axis and e is the eccentricity.
From Fig. 54, page 119, we have c = CO = CV  OV  aa(le) = ae.
An analogous result holds for the hyperbola [see Prdblem 5.73(c), page 139].
5.20. If a and c are as in Problem 5.19 and b is the length of the semiminor axis, prove
that (a) c = \Za 2  b\ (b) b = a^/l  A
(a) From Fig. 54, page 119, and the definition of an ellipse, we have
6 = QV = CZ__™ = «__^ or ys = ^=^ (i)
6 VE VE VE e
Also since the eccentricity is the distance from O to W divided by the distance from W
to the directrix AB [which is equal to CE], we have
OW/CE  e
or, using (1) and the result of Problem 5.19,
OW = eCE = e(CV + VE) = e[a + (ac)/e] = ea + ac = a
Then (OW) 2 = (OC) 2 + (CW) 2 or a 2 = b 2 + c 2 , i.e. c = V^ 2  & 2 
(6) From Problem 5.19 and part (a), a 2 = & 2 + a 2 e 2 or 6 = ay/1 e 2 .
KEPLER'S LAWS OF PLANETARY MOTION AND
NEWTON'S UNIVERSAL LAW OF GRAVITATION
5.21. Prove that if a planet is to revolve around the sun in an elliptical path with the sun
at a focus [Kepler's first law], then the central force necessary varies inversely as
the square of the distance of the planet from the sun.
If the path is an ellipse with the sun at a focus, then calling r the distance from the sun,
we have by Problem 5.16,
CHAP. 5] CENTRAL FORCES AND PLANETARY MOTION 129
* X ^costf (1)
1 + e cos o r p p
where e < 1. Then the central force is given as in Problem 5.7 by
f(l/u) = mh 2 u 2 (d 2 u/de 2 + u) = ^mh 2 u 2 /p (2)
on substituting the value of u in (1). From (2) we have on replacing u by 1/r,
f(r) = mh 2 /pr 2 = Kir 2 (S)
5.22. Discuss the connection of Newton's universal law of gravitation with Problem 5.21.
Historically, Newton arrived at the inverse square law of force for planets by using Kepler's
first law and the method of Problem 5.21. He was then led to the idea that perhaps all objects
of the universe were attracted to each other with a force which was inversely proportional to the
square of the distance r between them and directly proportional to the product of their masses.
This led to the fundamental postulate
_ GMm , Y v
where G is the universal gravitational constant. Equivalently, the law of force (3) of Problem 5.21
is the same as (1) where
K = GMm (2)
5.23. Prove Kepler's third law: The squares of the periods of the various planets are
proportional to the cubes of their corresponding semimajor axes.
If a and b are the lengths of the semimajor and semiminor axes, then the area of the
ellipse is nab. Since the areal velocity has magnitude h/2 [Problem 5.6], the time taken to sweep
over area vab, i.e. the period, is
p _ vab _ 2vab ,^\
h/2 ~ h K '
Now by Problem 5.17 equation (3), Problem 5.20(6), and Problem 5.13 equation (8), we have
6 = aVl ~ <?, p = a(le 2 ) = mhVK (2)
Then from (1) and (2) we find
P = 2irw>i 2 a^ 2 IK^ 2 or P 2 = 4ir 2 maVK
Thus the squares of the periods are proportional to the cubes of the semimajor axes.
5.24. Prove that GM = gR 2 .
On the earth's surface, i.e. r — R where R is the radius, the force of attraction of the earth
on an object of mass m is equal to the weight mg of the object. Thus if M is the mass of the
GMm/R 2 = mg or GM = gR 2
5.25. Calculate the mass of the earth.
From Problem 5.24, GM = gR 2 or M  gR 2 /G. Taking the radius of the earth as
6.38 X 10 8 cm, g  980 cm/sec 2 and G — 6.67 X 10~ 8 cgs units, we find M = 5.98 X 10 27 gra =
1.32 X 10 25 lb.
ATTRACTION OF OBJECTS
5.26. Find the force of attraction of a thin uniform rod of length 2a on a particle of
mass m placed at a distance b from its midpoint.
130
CENTRAL FORCES AND PLANETARY MOTION
[CHAP. 5
Choose the x axis along the rod and the y axis
perpendicular to the rod and passing through its
center O, as shown in Fig. 510. Let a be the mass
per unit length of the rod. The force of attraction
dF between an element of mass o dx of the rod and
m is, by Newton's universal law of gravitation,
dF
Gmadx
x 2 + b 2
Gmox dx
(sin e i — cos e j)
Gmvb dx
(X 2 + 62)3/2 ( X 2 + 62)3/2
Fig. 510
since from Fig. 510, sin e  x/y/x 2 + b 2 , cos $  b/y/x 2 + b 2 . Then the total force of attraction is
_, _ . C a Gmox dx _ . C a Gmabdx
~ l J ( x 2 +ft 2)3/2 3 J
=
(x 2 + b 2 )^ 2
_. C a Gmabdx
 2j X
(x 2 + ft 2 ) 3 ' 2
(x 2 + b 2 )*' 2
Xa
dx
(X 2 + 6 2 ) 3 /2
Let x = b tanfl in this integral. Then when x — 0, e — 0; and when x = a, 6 — tan 1 (a/6).
Thus the integral becomes
F =
2Gmabj I
tan 1 (a/b) b se( , 2 Q d$
2Gmcra
(b 2 sec 2 <s>) 3/ 2
o  ' by/a 2 +b 2
Since the mass of the rod is M = 2aa, this can also be written as
^, GMm
F = 3
by/a 2 +b 2
Thus we see that the force of attraction is directed from m to the center of the rod
and of magnitude 2Gmaa/by/a 2 + b 2 or GMmfby/a 2 + 6 2 .
5.27. A mass m lies on the perpendicular through the center of a uniform thin circular
plate of radius a and at distance b from the center. Find the force of attraction
between the plate and the mass m.
Method 1.
Let n be a unit vector drawn from point P where m
is located to the center O of the plate. Subdivide the
circular plate into circular rings [such as ABC in
Fig. 511] of radius r and thickness dr. If a is the mass
per unit area, then the mass of the ring is a{2wrdr).
S ince all points of the ring are at the same distance
y/r 2 + b 2 from P, the force of attraction of the ring on
m will be
,_ _ Ga{2vr dr)m
r 2 + 6 2
cos <f> n
Go 2irr dr mb
(1)
( r 2 4. £,2)3/2 "
where we have used the fact that due to symmetry the
resultant force of attraction is in the direction n. By
integrating over all rings from r = to r = a, we
find that the total attraction is
F = 2vGomb
r
r dr
( r 2 j. &2)3/2
b \Vr 2 + fe 2
Fig. 511
(2)
To evaluate the integral, let r 2 + 6 2
so that r dr = u du. Then since u = 6 when r =
and u = yja 2 + b 2 when r — a, the result is
CHAP. 5]
CENTRAL FORCES AND PLANETARY MOTION
131
IwGomb n I
v^+b* udu
2irGom n ( 1 —
u ° \ y/a 2 +b 2 ,
If we let a be the value of $ when r = a, this can be written
F = lirGom n (1  cos a) (3)
Thus the force is directed from m to the center O of the plate and has magnitude 2irG<rm(l — cos a).
Method 2.
The method of double integration can also be used. In such case the element of area at A is
rdrde where 6 is the angle measured from a line [taken as the x axis] in the plane of the
circular plate and passing through the center O. Then we have as in equation (1),
Ga(r dr do)mb
d¥
and by integrating over the circular plate
( r 2 ( J,2)3/2
F = Gamb
■/* S
r=0 ^0 =
r dr de
(r 2 + 6 2 ) 3/2
Xa
2irr dr
(r 2 + 6 2 )3/ 2
2TrGom n (1 — cos a)
5.28. A uniform plate has its boundary con
sisting of two concentric half circles of
inner and outer radii a and b respec
tively, as shown in Fig. 512. Find the
force of attraction of the plate on a
mass m located at the center 0.
It is convenient to use polar coordinates
(r, e). The element of area of the plate [shaded
in Fig. 512] is dA = r dr de, and the mass is
ardrde. Then the force of attraction between
dA and O is
_ _ G{ar dr de)m
Fig. 512
(cos e i + sin e j)
Thus the total force of attraction is
XC G{ardrde)m . . . . n ..
I —  — £ — — (cos e i + sin e j)
)=0 r=a
Gam In
W X=o
(cos o i + sine j) de = 2Gaw In (  ) j
Since M — o(±vb 2 — ^va 2 ), we have a = 2Mh(b 2 — a 2 ) and the force can be written
AGMm . fb\ .
* = ^2^2) ^ (^j,
The method of single integration can also be used by dividing the region between r = a and
r — b into circular rings as in Problem 5.27.
5.29. Find the force of attraction of a thin spherical shell of radius a on a particle P of
mass m at a distance r > a from its center.
Let O be the center of the sphere. Subdivide the surface of the sphere into circular elements
such as ABCDA of Fig. 513 below by using parallel planes perpendicular to OP.
The area of the surface element ABCDA as seen from Fig. 513 is
2^{a sin e)(a de) = 2va 2 sin e de
since the radius is a sin e [so that the perimeter is 2ir(a sin »)] and the thickness is ade. Then
if a is the mass per unit area, the mass of ABCDA is 2rra 2 a sin e de.
132
CENTRAL FORCES AND PLANETARY MOTION
[CHAP. 5
Since all points of ABCDA are at the same dis
tance w = AP from P, the force of attraction of
the element ABCDA on m is
7 „ G(2va 2 a sin 6 de)m
d¥ = s cos <p n
in*
(1)
where we have used the fact that from symmetry the
net force will be in the direction of the unit vector n
from P toward O. Now from Fig. 513,
PE
AP
PO  EO
AP
r — a cos e
w
(2)
Using (2) in (1) together with the fact that by the
cosine law
w 2 = a 2 + r 2 — 2ar cos B (3)
we find
dF
G{2ira 2 a sin e de)m{r — a cos e)
ade
(a 2 + r 2 — 2ar cos 0) 3/2
Then the total force is
F = 2TrGa 2 am
Fig. 513
"J.
(r — a cos e) sin e
(a 2 + r 2 — 2ar cos 0) 3/2
de
(•4)
We can evaluate the integral by using the variable w given by (3) in place of 6. When
= 0,
2ar + r 2 = (r — a) 2 so that w = r — a if r > a. Also when 6 = ir, w 2 —
a 2 + 2ar + r 2 = (r + a) 2 so that w = r + a. In addition, we have
2w dw = 2ar sin 6 de
' a 2 f r 2 _ w 2\ «)2
a COS
2ar
a 2 + r 2
Then (4) becomes
F =
irGaam n
x:
i +
dw —
2r
&TrGa 2 om n
r 2
5.30. Work Problem 5.29 if r < a.
In this case the force is also given by (4) of Problem 5.29. However, in evaluating the
integral we note that on making the substitution (3) of Problem 5.29 that e = yields w 2 = (a — r) 2
or w — a — r if r < a. Then the result (4) of Problem 5.29 becomes
F =
wGaom
Jo
1 
dw =
Thus there will be no force of attraction of a spherical shell on any mass placed inside. This
means that in such case a particle will be in equilibrium inside of the shell.
5.31. Prove that the force of attraction in Problem 5.29 is the same as if all the mass of
the spherical shell were concentrated at its center.
The mass of the shell is M = AttcPo. Thus the force is F = (GMm/r 2 )n, which proves the
required result.
5.32. (a) Find the force of attraction of a solid uniform sphere on a mass m placed outside
of it and (b) prove that the force is the same as if all the mass were concentrated
at its center.
(a) We can subdivide the solid sphere into thin concentric spherical shells. If p is the distance
of any of these shells from the center and dp is the thickness, then by Problem 5.29 the
force of attraction of this shell on the mass m is
CHAP. 5]
CENTRAL FORCES AND PLANETARY MOTION
133
_ G<x(4t7 P 2 d P )m
d¥ — • — s n
(1)
where a is the mass per unit volume. Then the total force obtained by integrating from
r = to r = a is
AirGomn f a „ .
^ dp
F =
Xa
> 2
G{% va s )am n
(*)
(6) Since the mass of the sphere is M — ^waPo, (2) can be written as F = (GMm/r 2 )n, which
shows that the force of attraction is the same as if all the mass were concentrated at
the center.
We can also use triple integration to obtain this result [see Problem 5.130].
5.33. Derive the result of Problems 5.29 and 5.30 by first finding the potential due to the
mass distribution.
The potential dV due to the element ABCDA is
G{2tto, 2 o sin e do)m __ G{2va 2 a sin e de)m
dV
Then the total potential is
— 2irGa 2 am
V« 2 + r 2 — 2ar cos 9
o V a2 + r 2 — 2ar cos
_ 2irGaam ^/ ( „ + y)2 ,/(„. r)2 }
If r > a this yields
If r < a it yields
4:irGa 2 cnn
GMm
r r
V = —kwGaom
Then if r > a the force is
F = VV = V
and if r < a the force is
GMm \
GMm
,2 l l
F = VV = V(4sGaam) =
in agreement with Problems 5.29 and 5.30.
MISCELLANEOUS PROBLEMS
5.34. An object is projected vertically upward from the earth's surface with initial
speed Vo. Neglecting air resistance, (a) find the speed at a distance H above the
earth's surface and (6) the smallest velocity of projection needed in order that the
object never return.
(a) Let r denote the radial distance of the object at time t
from the center of the earth, which we assume is fixed
[see Pig. 514]. If M is the mass of the earth and R is
its radius, then by Newton's universal law of gravitation
and Problem 5.29, the force between m and M is
F =
GMm
(1)
where rj is a unit vector directed radially outward from
the earth's center in the direction of motion of the object.
If v is the speed at time t, we have by Newton's sec
ond law,
Fig. 514
134 CENTRAL FORCES AND PLANETARY MOTION [CHAP. 5
tt)
Since the object starts from the earth's surface with speed v , we have v = v when r = R
so that <?! = vjj/2  GM/.R. Then (4) becomes
GMm
r 2 r i
dv
Tt
This can be written as
dv dr _
dr dt
GM
r 2
or
dv
dr
Then by integrating, we find
v 2 /2 =
GM/r + c t
2GM (ii) + ^
ve the earth'
Thus when the object is at height H above the earth's surface, i.e. r = R + H,
2GMH
(5)
R(R + H)
, 2 2GMH
V = ^ V °R(R + H)
Using Problem 5.24, this can be written
v = ^jvl
!~2 2gRH
V = \<~R+H
(6)
(b) As H * oo, the limiting speed (6) becomes
vX ~ 2GM/R or v^ 2 ,  2gR (7)
since lim /p „ = 1. The minimum initial speed occurs where (7) is zero or where
Ht» (it + /3)
v = V2GM/R = y/2gR (8)
This minimum speed is called the escape speed and the corresponding velocity is called the
escape velocity from the earth's surface.
5.35. Show that the magnitude of the escape velocity of an object from the earth's surface
is about 7 mi/sec.
From equation (*) of Problem 5.34, v = y/2gR. Taking g = 32 ft/sec 2 and R = 4000 mi,
we find v — 6.96 mi/sec.
5.36. Prove, by using vector methods primarily, that the path of a planet around the sun
is an ellipse with the sun at one focus.
Since the force F between the planet and sun is
F _ w dv _ GMm
¥ ~ m dt ~ r 2 ri {1)
, dv GM ,.
we have =r = xr t (2)
dt r 2
Also, by Problem 5.1, equation (4), we have
r X v = h (3)
dv av^ (£y
Now since r = rr., v = 37 — r—r + rr,. Thus from (S),
at at at
f dr, (j r \ dr,
h = rX v = „ lX (,— + „) = ^r.xJ W)
CHAP. 5] CENTRAL FORCES AND PLANETARY MOTION 135
From (2),
dv GM f dr i
[f *A drA or,
=  GM {("^><'.")«i = GM ^r
using equation (4) above and equation (7), page 5.
But since h is a constant vector, jr X h = — (v X h) so that
at at
d ^ r i
(vxh) = GM±
Integrating, v X h = GM r t + c
from which
r • (v X h) = GM t • r 1 + r • c = GMr + rr x • c = GMr + re cos e
where c is an arbitrary constant vector having magnitude c, and e is the angle between c and r x .
Since r • (v X h) = (r X v) • h = h • h = h 2 [see Problem 1.72(a), page 27],
h 2 = GMr + re cos 6
h 2 h 2 /GM
and so
GM + c cos e 1 + (c/GM) cos e
which is the equation of a conic. Since the only conic which is a closed curve is an ellipse, the
required result is proved.
5.37. Prove that the speed v of a particle moving in an elliptical path in an inverse square
field is given by
m \r a
where a is the semimajor axis.
By (8) of Problem 5.13, (4) of Problem 5.14 and (3) of Problem 5.17, we have
mh 2 n 2 . / 2Emh 2 \ ...
V = g^ = a(le 2 ) = a I K2~J (*)
from which E  K/2a (2)
Thus by the conservation of energy we have, using V = —Kir,
or v 2 = () (3)
m \r a I
Kf2 +h) (*)
We can similarly show that for a hyperbola,
m \r a
while for a parabola [which corresponds to letting a * °° in either (3) or (4)] ,
v 2 = 2K/mr
5.38. An artificial (manmade) satellite revolves about the earth at height H above the
surface. Determine the (a) orbital speed and (b) orbital period so that a man in the
satellite will be in a state of weightlessness.
(a) Assume that the earth is spherical and has radius R. Weightlessness will result when the
centrifugal force [equal and opposite to the centripetal force, i.e. the force due to the cen
136
CENTRAL FORCES AND PLANETARY MOTION
[CHAP. 5
tripetal acceleration] acting on the man due to rotation of the satellite just balances his
attraction to the earth. Then if v is the orbital speed,
mv l _ GMm _ gR*m _ R
R + H ~ (R + H)* ~ (R + H)* or v ° ~ R + H
If H is small compared with R, this is y/Rg approximately.
y/(R + H)g
(b)
Orbital speed
distance traveled in one revolution
time for one revolution, or period
Thus
Then from part (a)
_ 2tt(R + H)
_ 2u(R + H) _
r — — ^ 7
«0
R + H
R
: )V 1 ?
If H is small compared with R, this is 2iry/R/g approximately.
5.39. Calculate the (a) orbital speed and (b) period in Problem 5.38 assuming that the
height H above the earth's surface is small compared with the earth's radius.
Taking the earth's radius as 4000 miles and g = 32 ft/sec 2 , we find (a) v = y/Rg 
4.92 mi/sec and (b) P = 2n^R/g  1.42 hr = 85 minutes, approximately.
5.40. Find the force of attraction of a solid sphere of radius a on a particle of mass m at
a distance b < a from its center.
By Problem 5.30 the force of attraction of any spherical
shell containing m in its interior [such as the spherical shell
shown dashed in Fig. 515] is zero.
Thus the force of attraction on m is the force due to a
sphere of radius b < a with center at O. If a is the mass
per unit volume, the force of attraction is
G(f7r&3)<rm/&2 = (^Gam)b
Thus the force varies as the distance 6 from the mass to the
center. Fig. 515
Supplementary Problems
CENTRAL FORCES AND EQUATIONS OF MOTION
5.41. Indicate which of the following central force fields are attractive toward origin O and which
are repulsive from O. (a) F = 4^; (6) F = Kxjyfr, K > 0; (c) F = r(r ljr^r 2 + 1);
(d) F = sin wr r v
Ans. (a) attractive; (6) repulsive; (<;) attractive if < r < 1, repulsive if r > 1; (d) repulsive for
2n < r < 2n + 1, attractive for 2n+l<r<2n + 2 where n — 0, 1, 2, 3 . . . .
5.42. Prove that in rectangular coordinates the magnitude of the areal velocity is ^{xy — yx).
5.43. Give an example of a force field directed toward a fixed point which is not a central force field.
CHAP. 5] CENTRAL FORCES AND PLANETARY MOTION 137
5.44. Derive equation (7), page 117.
5.45. If a particle moves in a circular orbit under the influence of a central force at its center, prove that
its speed around the orbit must be constant.
5.46. A particle of mass m moves in a force field defined by F = Kr^r 3 . If it starts on the positive
x axis at distance a away from the origin and moves with speed v in direction making angle
a with the positive x axis, prove that the differential equation for the radial position r of the
particle at any time Ms „ „
d2 r _ (K — ma 2 v\ sin 2 a)
dt 2 mr 3
5.47. (a) Show that the differential equation for the orbit in Problem 5.46 is given in terms of u = 1/r by
+ (1 — y)u — where y —
do 2 ' ma 2 v 2 sin 2 a
(b) Solve the differential equation in (a) and interpret physically.
5.48. A particle is to move under the influence of a central force field so that its orbital speed is
always constant and equal to v . Determine all possible orbits.
POTENTIAL ENERGY AND CONSERVATION OF ENERGY
5.49. Find the potential energy or potential corresponding to the central force fields defined by
(a) F = Krjr*, (b) F = {air 2 + 0/r»)r lf (c) F = Krr u (d) F = r x /yfF, (e) F = sin ar r^
Ans. (a) K/2r 2 , (b) a/r + p/2r 2 , (c) \Kr 2 , (d) 2ypr, (e) (cos»r)/ir
5.50. (a) Find the potential energy for a particle which moves in the force field F = —Krjr 2 . (b) How
much work is done by the force field in (a) in moving the particle from a point on the circle
r = a > to another point on the circle r = 6 > 0? Does the work depend on the path? Explain.
Ans. (a) Kir, (6) K(ab)/ab
5.51. Work Problem 5.50 for the force field F = Krjr. Ans. (a) K\nr, (6) K In (a/6)
5.52. A particle of mass m moves in a central force field defined by F = —Krjr*. (a) Write an equation
for the conservation of energy. (6) Prove that if E is the total energy supplied to the particle,
then its speed is given by v = yjKlmr 2 + 2E/m.
5.53. A particle moves in a central force field defined by F = —Kr 2 r v It starts from rest at a point
o n the circle r = a. (a) Prove that when it reaches the circle r = b its speed will be
V2#(a 3  6 3 )/3w and that (6) the speed will be independent of the path.
5.54. A particle of mass m moves in a central force field F = Kxjr n where K and n are constants.
It starts from rest at r = a and arrives at r = with finite speed v . (a) Prove that we must
have n < 1 and K > 0. (b) Prove that v = y/2Ka i  n /m{n 1). (c) Discuss the physical sig
nificance of the results in (a).
5.55. By differentiating both sides of equation (IS), page 117, obtain equation (6).
DETERMINATION OF ORBIT FROM CENTRAL FORCE OR
CENTRAL FORCE FROM ORBIT
5.56. A particle of mass m moves in a central force field given in magnitude by f(r) = —Kr where
K is a positive constant. If the particle starts at r — a, e = with a speed v in a direction
perpendicular to the x axis, determine its orbit. What type of curve is described?
5.57. (a) Work Problem 5.56 if the speed is v in a direction making angle a with the positive * axis.
(6) Discuss the cases a — 0, a = ir and give the physical significance.
138 CENTRAL FORCES AND PLANETARY MOTION [CHAP. 5
5.58. A particle moving in a central force field located at r = describes the spiral r = e~ . Prove
that the magnitude of the force is inversely proportional to r 3 .
5.59. Find the central force necessary to make a particle
describe the lemniscate r 2 = a 2 cos 2d [see Fig. 516].
Ans. A force proportional to r~ 7 . r 2 = a*cos2»
5.60. Obtain the orbit for the particle of Problem 5.46 and
describe physically.
5.61. Prove that the orbits r — e~ e and r = 1/8 are both
possible for the case of an inverse cube field of force.
Explain physically how this is possible. Fig. 516
5.62. (a) Show that if the law of force is given by
F =
r 4 cos e r 2 cos 3 e
then a particle can move in the circular orbit r — 2a cos 8. (b) What can you conclude about
the uniqueness of forces when the orbit is specified? (c) Answer part (b) when the forces are
central forces.
5.63. (a) What central force at the origin O is needed to make a particle move around O with a speed
which is inversely proportional to the distance from O. (b) What types of orbits are possible in
such case? Ans. (a) Inverse cube force.
5.64. Discuss the motion of a particle moving in a central force field given by F = (a/r 2 + )8/r 3 )r 1 .
5.65. Prove that there is no central force which will enable a particle to move in a straight line.
5.66. Complete the integration of equation (12) of Problem 5.13, page 125 and thus arrive at equation
(5) of the same problem. [Hint. Let r = 1/u.]
5.67. Suppose that the orbit of a particle moving in a central force field is given by 8 = e(r). Prove
mh 2 [2e' + re" + r 2 (e') 3 ]
that the law of force is 5 , ,. 3 where primes denote differentiations with
respect to r.
5.68. (a) Use Problem 5.67 to show that if 6 = 1/r, the central force is one of attraction and varies
inversely as r 3 . (6) Graph the orbit in (a) and explain physically.
CONIC SECTIONS. ELLIPSE, PARABOLA AND HYPERBOLA
12
5.69. The equation of a conic is r = ^—, • Graph the conic, finding (a) the foci, (6) the vertices,
S + cos 8
(c) the length of the major axis, (d) the length of the minor axis, (e) the distance from the center
to the directrix.
24
5.70. Work Problem 5.69 for the conic r =
3 + 5 cos 8 '
5.71. Show that the equation of a parabola can be written as r — p sec 2 (8/2).
5.72. Find an equation for an ellipse which has one focus at the origin, its center at the point (—4, 0),
and its major axis of length 10. Ans. r = 9/(5 + 4 cos 8)
CHAP. 5] CENTRAL FORCES AND PLANETARY MOTION 139
5.73. In Fig. 517, SR or TN is called the minor axis of %\ x/'
the hyperbola and its length is generally denoted >^ //
by 26. The length of the major axis VU is 2a, \ x r re's
while the distance between the foci O and O' is 2c \l X \ / / \f
[i.e. the distance from the center C to a focus O m _\ n ., x y
or O' is C]. °/ y / /Cx U \ '
(a) Prove that c 2 = a 2 + b 2 . /y y _ ^H\
(b) Prove that b = aVe 2 — 1 where e is the eccen // \
tricity. >^ x \
(c) Prove that c = etc. Compare with results for y ^>
the ellipse. Fig. 517
5.74. Derive equation (22), page 119, for a hyperbola.
5.75. In rectangular coordinates the equations for an ellipse and hyperbola in standard form are given by
* 2 ,y 2 1 , x 2 v 2
tf + V 2 = 1 and V 2 ~V = x
respectively, where a and b are the lengths of the semimajor and semiminor axes. Graph these
equations, locating vertices, foci and directrices, and explain the relation of these equations to
equations [19), page 118, and (22), page 119.
5.76. Using the alternative definitions for an ellipse and hyperbola given on pages 118119, obtain the
equations (19) and (22).
5.77. Prove that the angle between the asymptotes of a hyperbola is 2 cos 1 (1/e).
KEPLER'S LAWS AND NEWTON'S LAW OF GRAVITATION
5.78. Assuming that the planet Mars has a period about the sun equal to 687 earth days approximately,
find the mean distance of Mars from the sun. Take the distance of the earth from the sun as
93 million miles. Ans. 140 million miles
5.79. Work Problem 5.78 for (a) Jupiter and (b) Venus which have periods of 4333 earth days and
225 earth days respectively. Ans. (a) 484 million miles, (b) 67 million miles
5.80. Suppose that a small spherical planet has a radius of 10 km and a mean density of 5 gm/cm 3 .
(a) What would be the acceleration due to gravity at its surface? (6) What would a man weigh
on this planet if he weighed 80 kgwt on earth?
5.81. If the acceleration due to gravity on the surface of a spherically shaped planet P is g P while its
mean density and radius are given by a P and R P respectively, prove that g P = %irGR P e P where G
is the universal gravitational constant.
5.82. If L, M, T represent the dimensions of length, mass and time, find the dimensions of the universal
gravitational constant. Ans. L 3 M _1 T~ 2
5.83. Calculate the mass of the sun using the fact that the earth is approximately 150 X 10 6 kilometers
from it and makes one complete revolution about it in approximately 365 days. Ans. 2 X 10 30 kg
5.84. Calculate the force between the sun and the earth if the distance between the earth and the sun is
taken as 150 X 10 6 kilometers and the masses of the earth and sun are 6 X 10 24 kg and 2 X 10 30 kg
respectively. Ans. 1.16 X 10 24 newtons
ATTRACTION OF OBJECTS
5.85. Find the force of attraction of a thin uniform rod of length a on a mass m outside the rod but on
the same line as the rod and distance 6 from an end. Ans. GMm/b(a + b)
5.86. In Problem 5.85 determine where the mass of the rod should be concentrated so as to give the
same force of attraction. Ans. At a point in the rod a distance y/b(a + b) — 6 from the end
14 () CENTRAL FORCES AND PLANETARY MOTION [CHAP. 5
5.87. Find the force of attraction of an infinitely long thin uniform rod on a mass m at distance b
from it. Arts. Magnitude is 2Gmafb
5.88. A uniform wire is in the form of an arc of a circle of radius 6 and central angle ^. Prove that
the force of attraction of the wire on a mass m placed at the center of the circle is given in
magnitude by 2Gam sin (v fr/2)
6V ° r *
where M is the mass of the wire and a is the mass per unit length. Discuss the cases ^ = v/2
and yp — 7r.
5.89. In Fig. 518, AB is a thin rod of length 2a and m
is a mass located at point C a distance 6 from the
rod. Prove that the force of attraction of the rod .
on m has magnitude ^^ c
— — sin Ua + (3) D/
ao ^\
/^ \ 6 \
in a direction making an angle with the rod s' \ \
given by /^ v \^^ E ^A~
_ 1 / cos/3 + cosoA ^ r ~~ ~~' '■■""" ^ /?
tan VsiniSsinay I 2a I
Discuss the case a = /? and compare with Prob
lem 5.26. Fig. 518
5.90. By comparing Problem 5.89 with Problem 5.88, prove that the rod of Problem 5.89 can be
replaced by a wire in the form of circular arc BEG [shown dashed in Fig. 518] which has its
center at C and is tangent to the rod at E. Prove that the direction of the attraction is toward
the midpoint of this arc.
5.91. A hemisphere of mass M and radius a has a particle of mass m located at its center. Find the
force of attraction if (a) the hemisphere is a thin shell, (6) the hemisphere is solid.
Ana. (a) GMm/2a 2 , (b) 36 Mm/2a 2
5.92. Work Problem 5.91 if the hemisphere is a shell having outer radius a and inner radius b.
5.93. Deduce from Kepler's laws that if the force of attraction between sun and planets is given in
magnitude by ym/r 2 , then y must be independent of the particular planet.
5 94 A cone has height H and radius a. Prove that the force of attraction on a particle of mass m
■. ^ §GMm A H \
placed at its vertex has magnitude — ^ — 1 1 ~ / 24 W2 / "
5.95. Find the force of attraction between two nonintersecting spheres.
5 96 A particle of mass m is placed outside of a uniform solid hemisphere of radius a at a distance a
on a line perpendicular to the base through its center. Prove that the force of attraction is
given in magnitude by GMm{\f2 — l)/a 2 .
5.97. Work (a) Problem 5.26, (b) Problem 5.27, and (c) Problem 5.94 by first finding the potential.
MISCELLANEOUS PROBLEMS
5.98. A particle is projected vertically upward from the earth's surface with initial speed v .
(a) Prove that the maximum height H reached above the earth's surface is H = v 2 R/(2gR  v\).
(b) Discuss the significance of the case where v\ = 2gR.
(c) Prove that if H is small, then it is equal to v%/2g very nearly.
CHAP. 5] CENTRAL FORCES AND PLANETARY MOTION 141
5.99. (a) Prove that the time taken to reach the maximum height of Problem 5.98 is
J R + H J [H
^~2tUr
~ R + H _JRH
2R wo \R + H t
(b) Prove that if H is very small compared with R, then the time in (a) is very nearly ^2H/g.
5.100. (a) Prove that if an object is dropped to the earth's surface from a height H, then if air
resistance is negligible it will hit the earth with a speed v  ^2gRH/(R + H) where R is
the radius of the earth.
(b) Calculate the speed in part (a) for the cases where H = 100 miles and H = 10,000 miles
respectively. Take the radius of the earth as 4000 miles.
5.101. Find the time taken for the object of Problem 5.100 to reach the earth's surface in each of
the two cases.
5.102. What must be the law of force if the speed of a particle in a central force field is to be
proportional to r~ n where n is a constant?
5.103. What velocity must a space ship have in order to keep it in an orbit around the earth at a
distance of (a) 200 miles, (6) 2000 miles above the earth's surface"
>9
5.104. An object is thrown upward from the earth's surface with velocity v . Assuming that it returns
to earth and that air resistance is negligible, find its velocity on returning.
5.105. (a) What is the work done by a space ship of mass m in moving from a distance a above the
earth's surface to a distance 6?
(6) Does the work depend on the path? Explain. Ans. (a) GmM(a — b)/ab
5.106. (o) Prove that it is possible for a particle to move in a circle of radius a in any central force
field whose law of force is /(r).
(b) Suppose the particle of part (a) is displaced slightly from its circular orbit. Prove that
it will return to the orbit, i.e. the motion is stable, if
w 4U .l • af ' {a) + 3/(a) > °
but is unstable otherwise.
(c) Illustrate the result in (6) by considering f(r) = 1/r" and deciding for which values of n
stability can occur. Ans. (c) For n < 3 there is stability.
5.107. If the moon were suddenly stopped in its orbit, how long would it take to fall to the earth
assuming that the earth remained at rest? Ans. About 4 days 18 hours
5.108. If the earth were suddenly stopped in its orbit, how long would it take for it to fall into the sun?
Ans. About 65 days
5.109. Work Problem 5.34, page 133, by using energy methods.
5.110. Find the velocity of escape for an object on the surface of the moon. Use the fact that the
acceleration due to gravity on the moon's surface is approximately 1/6 that on the earth and
that the radius of the moon is approximately 1/4 of the earth's radius. Ans. 1.5 mi/sec
5.111. An object is dropped through a hole bored through the center of the earth. Assuming that the
resistance to motion is negligible, show that the speed of the particle as it passes through the
center of the earth is slightly less than 5 mi/sec.
[Hint. Use Problem 5.40, page 136.]
5.112. In Problem 5.111 show that the time taken for the object to return is about 85 minutes.
14 2 CENTRAL FORCES AND PLANETARY MOTION [CHAP. 5
5.113. Work Problems 5.111 and 5.112 if the hole is straight but does not pass through the center of
the earth.
5.114. Discuss the relationship between the results of Problems 5.111 and 5.112 and that of Problem 5.39.
5.115. How would you explain the fact that the earth has an atmosphere while the moon has none?
5.116. Prove Theorem 5.1, page 120.
5.117. Discuss Theorem 5.1 if the spheres intersect.
5.118. Explain how you could use the result of Problem 5.27 to find the force of attraction of a solid
sphere on a particle.
5.119. Find the force of attraction between a uniform circular ring of outer radius a and inner radius 6
and a mass m located on its axis at a distance 6 from its center.
5.120. Two space ships move about the earth on the same elliptical path of eccentricity e. If they are
separated by a small distance D at perigee, prove that at apogee they will be separated by the
distance D(l  e)/(l + c).
5.121. (a) Explain how you could calculate the velocity of escape from a planet. (6) Use your method to
calculate the velocity of escape from Mars. Ans. (6) 5 km/sec, or about 3 mi/sec
5.122. Work Problem 5.121 for (a) Jupiter, (b) Venus. Ans. (a) about 38 mi/sec, (6) about 6.3 mi/sec
5.123. Three infinitely long thin uniform rods having the same mass per unit length lie in the same plane
and form a triangle. Prove that force of attraction on a particle will be zero if and only if the
particle is located at the intersection of the medians of the triangle.
5.124. Find the force of attraction between a uniform rod of length a and a sphere of radius b if they
do not intersect and the line of the rod passes through the center.
5.125. Work Problem 5.124 if the rod is situated so that a line drawn from the center perpendicular to the
line of the rod bisects the rod.
5.126. A satellite of radius a revolves in a circular orbit about a planet of radius b with period P.
If the shortest distance between their surfaces is c, prove that the mass of the planet is
4^2(a + b + c)3/GP 2 .
5.127. Given that the moon is approximately 240,000 miles from the earth and makes one complete
revolution about the earth in 27£ days approximately, find the mass of the earth.
Ans. 6 X 10 24 kg
5.128. Discuss the relationship of Problem 5.126 with Kepler's third law.
5.129. Prove that the only central force field F whose divergence is zero is an inverse square force field.
5.130. Work Problem 5.32, page 132, by using triple integration.
5.131. A uniform solid right circular cylinder has radius a and height H. A particle of mass m is placed
on the extended axis of the cylinder so that it is at a distance D from one end. Prove that the
force of attraction is directed along the axis and given in magnitude by
^^{H + Va2 + Z)2 Va2 + (D + H)2}
a 2 H
5.132. Suppose that the cylinder of Problem 5.131 has a given volume. Prove that the force of attraction
when the particle is at the center of one end of the cylinder is a maximum when alH = £(9  y/ll).
5.133. Work (a) Problem 5.26 and (&) Problem 5.27 assuming an inverse cube law of attraction.
CHAP. 5] CENTRAL FORCES AND PLANETARY MOTION 143
5.134. Do the results of Problems 5.29 and 5.30 apply if there is an inverse cube law of attraction?
Explain.
5.135. What would be the velocity of escape from the small planet of Problem 5.80?
5.136. A spherical shell of inner radius a and outer radius 6 has constant density a. Prove that the
gravitational potential V(r) at distance r from the center is given by
V(r) =
2ua(b 2 a 2 ) r < a
27r(j(b 2  lr 2 )  4™a 3 /3r a < r < b
47tct(6 3 — a 3 )/3r r > b
5.137. If Einstein's theory of relativity is taken into account, the differential equation for the orbit
of a planet becomes
d 2 u , K , .
do 2 mh 2 '
where y = 3K/mc 2 , c being the speed of light, (a) Prove that if axes are suitably chosen, then
the position r of the planet can be determined approximately from
mh 2 /K , 1 „, L9
r ■=  — ; where a — 1 — yK/mh 2
1 + e cos aO
(b) Use (a) to show that a planet actually moves in an elliptical path but that this ellipse slowly
rotates in space, the rate of angular rotation being 2iryK/mh 2 . (c) Show that in the case of
Mercury this rotation amounts to 43 seconds of arc per century. This was actually observed, thus
offering experimental proof of the validity of the theory of relativity.
5.138. Find the position of a planet in its orbit around the sun as a function of time t measured from
where it is furthest from the sun.
5.139. At apogee of 200 miles from the earth's surface, two space ships in the same elliptical path are
500 feet apart. How far apart will they be at perigee 150 miles assuming that they drift without
altering their path in any way?
5.140. A particle of mass m is located on a perpendicular line through the center of a rectangular plate
of sides 2a and 26 at a distance D from this center. Prove that the force of attraction of the plate
on the particle is given in magnitude by
GMm . _j / ab
ab Sin \^(a 2 + D 2 )(b 2 + D 2 )
5.141. Find the force of attraction of a uniform infinite plate of negligible thickness and density a
on a particle at distance D from it. Ans. 2iroGm
5.142. Points where r = are called apsides [singular, apsis], (a) Prove that apsides for a central
force field with potential V(r) and total energy E are roots of the equation V(r) + h 2 /2r 2 = E.
(b) Find the apsides corresponding to an inverse square field of force, showing that there are
two, one or none according as the orbit is an ellipse, hyperbola or parabola.
5.143. A particle moving in a central force field travels in a path which is the cycloid r = a(lcose).
Find the law of force. Ans. Inverse fourth power of r.
5.144. Set up equations for the motion of a particle in a central force field if it takes place in a medium
where the resistance is proportional to the instantaneous speed of the particle.
5.145. A satellite has its largest and smallest orbital speeds given by v max and i; min respectively. Prove
that the eccentricity of the orbit in which the satellite moves is equal to
max ' v r
5.146. Prove that if the satellite of ProbLem 5.145 has a period equal to t, then it moves in an elliptical
path having major axis whose length is  y/v max v min .
Chapter 6
MOVING
COORDINATE SYSTEMS
NONINERTIAL COORDINATE SYSTEMS
In preceding chapters the coordinate systems used to describe the motions of particles
were assumed to be inertial [see page 33]. In many instances of practical importance,
however, this assumption is not warranted. For example, a coordinate system fixed in
the earth is not an inertial system since the earth itself is rotating in space. Consequently
if we use this coordinate system to describe the motion of a particle relative to the earth
we obtain results which may be in error. We are led therefore to consider the motion of
particles relative to moving coordinate systems.
ROTATING COORDINATE SYSTEMS
In Fig. 61 let XYZ denote an inertial coordinate
system with origin O which we shall consider fixed
in space. Let the coordinate system xyz having the
same origin O be rotating with respect to the XYZ
system.
Consider a vector A which is changing with
time. To an observer fixed relative to the xyz system
the time rate of change of A = Aii + A 2 j + A 3 k is
found to be
dA
dt
dAi
~df
i +
dA,
dt
i +
dAt
dt
(1)
where subscript M indicates the derivative in the
moving (xyz) system.
However, the time rate of change of A relative
to the fixed XYZ system symbolized by the subscript
F is found to be [see Problem 6.1]
Fig. 61
dA
dt
dA
dt
+ o> x A
(2)
where a> is called the angular velocity of the xyz system with respect to the XYZ system.
DERIVATIVE OPERATORS
Let D F and D M represent time derivative operators in the fixed and moving systems.
Then we can write the operator equivalence
D F = D M + oX (3)
This result is useful in relating higher order time derivatives in the fixed and moving
systems. See Problem 6.6.
144
CHAP. 6] MOVING COORDINATE SYSTEMS 145
VELOCITY IN A MOVING SYSTEM
If, in particular, vector A is the position vector r of a particle, then (2) gives
+ *>Xr a)
dr _ dr
dt f ~ dt
or D F r = D M x + <o X r (5)
Let us write
v pF = dr/dt \ F = D F r = velocity of particle P relative to fixed system
v p(M = dr/dt \ M = D M r = velocity of particle P relative to moving system
v mif = • x r = velocity of moving system relative to fixed system.
Then (4) or (5) can be written
v pif = v PM + v MF (6)
ACCELERATION IN A MOVING SYSTEM
If D 2 F = d 2 /dt 2 \ F and D^ = d 2 /dt 2  M are second derivative operators with respect to t
in the fixed and moving systems, then application of (3) yields [see Problem 6.6]
D 2 F r = D 2 M r + (Z> M <o) X r + 2« X D M r + • X (• x r) (7)
Let us write
a PF = d 2 r/dt 2 \ F = D F r = acceleration of particle P relative to fixed system
a PM = d 2 r/dt 2 \ M = D^r = acceleration of particle P relative to moving system
a Mir = ( D m<*) X r + 2a. X D M r + « X (o, X r)
= acceleration of moving system relative to fixed system
Then (7) can be written
*PF **PM
"*" a MF (#)
X r (11)
M/
CORIOLIS AND CENTRIPETAL ACCELERATION
The last two terms on the right of (7) are called the Coriolis acceleration and centripetal
acceleration respectively, i.e.,
Coriolis acceleration = 2o» x D M r — 2<o x v M (9)
Centripetal acceleration = • x («• x r) (10)
The second term on the right of (7) is sometimes called the linear acceleration, i.e.,
/j
Linear acceleration = (D^a) x r = (~
v M ' \dt
and Z> M o> is called the angular acceleration. For many cases of practical importance [e.g. in
the rotation of the earth] <a is constant and D M <a = 0.
The quantity — <o x (o> x r) is often called the centrifugal acceleration.
MOTION OF A PARTICLE RELATIVE TO THE EARTH
Newton's second law is strictly applicable only to inertial systems. However, by using
(7) we obtain a result valid for noninertial systems. This has the form
mD^r = F  m(D M a) X r  2m(u x D M r)  m«x( B Xr) (12)
where F is the resultant of all forces acting on the particle as seen by the observer in the
fixed or inertial system.
146
MOVING COORDINATE SYSTEMS
[CHAP. 6
In practice we are interested in expressing the equations of motion in terms of quantities
as determined by an observer fixed on the earth [or other moving system]. In such case
we may omit the subscript M and write (12) as
d 2 v
m
dt 2
m(m X r) — 2m(o> x v) — m[» x (© X r)]
(13)
For the case of the earth rotating with constant angular o> about its axis, *> = and
(13) becomes , 2
m
dt 2
2m(<o X v)  m[» X (o> x r)]
(U)
CORIOLIS AND CENTRIPETAL FORCE
Referring to equations (13) or (14) we often use the following terminology
Coriolis force = 2ra(« x r) = 2ra(o> x v)
Centripetal force = m[« x (» x r)]
Centrifugal force = m[» x (» x r)]
MOVING COORDINATE SYSTEMS IN GENERAL
In the above results we assumed that the coordi
nate systems xyz and XYZ [see Fig. 61] have com
mon origin O. In case they do not have a common
origin, results are easily obtained from those already
considered.
Suppose that R is the position vector of origin Q
relative to origin [see Fig. 62]. Then if R and
R denote the velocity and acceleration of Q relative
to O, equations (5) and (7) are replaced respectively
by
D F r
= R + ZX„r + <oXr
= R +
dr
dt
+ <» xr
(15)
and
Fig. 62
Dl* = R + D 2 M r + (D M <o) x r + 2. x D M r + . x („ x r)
= R +S+ iXr + 2«Xv + «)X( a Xr)
Similarly equation (14) is replaced by
m
^r
dt 2
= F  2m(u X v)  m[<a x(<aX r)]  mR
(16)
(17)
THE FOUCAULT PENDULUM
Consider a simple pendulum consisting of a long string and heavy bob suspended
vertically from a f rictionless support. Suppose that the bob is displaced from its equilibrium
position and is free to rotate in any vertical plane. Then due to the rotation of the earth,
the plane in which the pendulum swings will gradually precess about a vertical axis. In the
northern hemisphere this precession is in the clockwise direction if we look down at the
earth's surface. In the southern hemisphere the precession would be in the counterclock
wise direction.
Such a pendulum used for detecting the earth's rotation was first employed by Foucault
in 1851 and is called Foucault's pendulum.
CHAP. 6]
MOVING COORDINATE SYSTEMS
147
Solved Problems
ROTATING COORDINATE SYSTEMS
6.1. An observer stationed at a point which is fixed relative to an xyz coordinate system
with origin O [see Fig. 61, page 144] observes a vector A = Ad + A 2 j + Ask. and
calculates its time derivative to be —fri + ~rfj~i + ~dT^ Later, he finds that he
and his coordinate system are actually rotating with respect to an XYZ coordinate
system taken as fixed in space and having origin also at 0. He asks, "What would
be the time derivative of A for an observer who is fixed relative to the XYZ coordi
nate system?"
If
dA
dt
j dA
and —rr
f dt
denote respectively the time derivatives of A relative to the
fixed and moving systems, show that there exists a vector quantity <*> such that
dA
dt
dA
dt
+ a X A
To the fixed observer the unit vectors i, j,k actually change with time. Hence such an
observer would compute the time derivative as
dA
dt
i.e.,
dA x
dt
dAI
dt \f
dA
2 .
dt
dA 3 ^
dt
i + ^i + ^k + A^ + A 2 f t + A ZTt
di
dt
d\
dk
dAl
dt \m
, di . d) a dk
+ ^1^7 + A 2^ + A 3 —
dt
dt
dt
(1)
(2)
Since i is a unit vector, di/dt is perpendicular to i and must therefore lie in the plane
of j and k. Then
di/dt = « J + a 2 k (S)
Similarly,
dj/dt = a 3 k + a 4 i
dk/dt = a 5 i + a 6 j
. dj ,di . _
U)
>dy
di
From i«j = 0, differentiation yields i * jT+ ji* j = 0 But i» 3i—a 4 from (4) and Tz'j = a l
from (3). Thus a 4 = — a v
dt dt
dt
dt
dk , di
dt dt
Similarly from i«k = 0, i»^ + ^»k = and a 5 = —a 2 ; from j»k = 0, J'^t+Tt* k =
and a 6 = — a 3 . Then
.rfk+dl.i, 
dt dt
di/dt = aj + a 2 k, dj/dt = a 3 k — a t i, dk/dt = —a 2 i — atf
It follows that
. di
^dt
which can be written as
+ A 2 ^ + A s ^ = («iA 2  a 2 A 3 )i + («iA l a s A s )j + (a 2 A! + a 3 A 2 )k (6)
dt
i
j
k
«3
_ «2
<*i
Ai
A 2
A 3
Then if we choose a 3 = « lf — a 2 = <o 2 , a t = w 3 this determinant becomes
i J k
«! W 2 <0 3
= <■> X A
where <a = u 1 i + « 2 j + w 3 k.
148
MOVING COORDINATE SYSTEMS
[CHAP. 6
From (2) and (6) we find, as required,
dA dA
dt j? dt
+ w X A
The vector quantity o» is the angular velocity of the moving system relative to the fixed system.
6.2. Let D F and D M be symbolic time derivative operators in the fixed and moving systems
respectively. Demonstrate the operator equivalence
By definition D F A
dA
dt
DmA = ~di
D F = D M + «x
= derivative in fixed system
= derivative in moving system
Then from Problem 6.1,
D F A = D M A + »XA = (D M + co X)A
which shows the equivalence of the operators D F = D M + o> X .
6.3. Prove that the angular acceleration is the same in both XYZ and xyz coordinate
systems.
Let A = <o in Problem 6.1. Then
da
~dt
da
~dl
+ to X
da
~di
Since dm/dt is the angular acceleration, the required statement is proved.
VELOCITY AND ACCELERATION IN MOVING SYSTEMS
6.4. Determine the velocity of a moving particle as seen by the two observers in
Problem 6.1.
Replacing A by the position vector r of the particle, we have
dr
dt
dr
dt
+ o, X r
(1)
If r is expressed in terms of the unit vectors i, j,k of the moving coordinate system, then the
velocity of the particle relative to this system is, on dropping the subscript M,
dr
dt
dx . dy^ . dz ,
dt di 3 dt
and the velocity of the particle relative to the fixed system is from (1)
dr
~di
dr . v
dt + wXr
(2)
(3)
The velocity (3) is sometimes called the true velocity, while (2) is the apparent velocity.
6.5. An xyz coordinate system is rotating with respect to an XYZ coordinate system
having the same origin and assumed to be fixed in space [i.e. it is an inertial system].
The angular velocity of the xyz system relative to the XYZ system is given by
a) = 2ti  t 2 j + (2t + 4)k where t is the time. The position vector of a particle at
time t as observed in the xyz system is given by r = (t 2 + l)i  6ij + 4£ 3 k. Find
(a) the apparent velocity and (b) the true velocity at time t = 1.
CHAP. 6] MOVING COORDINATE SYSTEMS 149
(a) The apparent velocity at any time t is
dr/dt = 2ti  6j + 12t 2 k
At time t = 1 this is 2i  6j + 12k.
(6) The true velocity at any time t is
dr/dt I.Xr = (2ti  6j + 12t%) + [2ti  t 2 ) + (2t + 4)k] X [(t 2 + l)i  6£j + 4£%]
At time £ = 1 this is
2i  6j + 12k +
i J k
216
26 4
34i  2j + 2k
6.6. Determine the acceleration of a moving particle as seen by the two observers in
Problem 6.1.
The acceleration of the particle as seen by the observer in the fixed XYZ system is
D F x — D F {D F r). Using the operator equivalence established in Problem 6.2, we have
D F (D F r) = D F (D M r + *> X r)
= (D M + «» X )(D M r + a. x r)
= D M {D M r + » X r) + u X (D M r + .Xr)
= D^r + D M (a> X r) + o> X D M r + «* X («• X r)
or since D M (<* X r) = (D M m) X r + » X (D M r),
Dlr  DmT + (D MW ) X r + 2.X (D M r) + • X (• X r) (1)
If r is the position vector expressed in terms of i, j,k of the moving coordinate system, then
the acceleration of the particle relative to this system is, on dropping the subscript M,
<ffr _ d?x. d 2 y. d*z
dt 2 dt 2 * + dt 2 * dt 2 (g)
The acceleration of the particle relative to the fixed system is given from (1) as
d 2 r I (Pr , d«„ , „ .. /dr
f
The acceleration (3) is sometimes called the true acceleration, while (2) is the apparent acceleration
dt 2 \ F ^ + ^ Xr + 2wX U^ + ttX(wXr) w
6.7. Find (a) the apparent acceleration and (b) the true acceleration of the particle in
Problem 6.5.
(a) The apparent acceleration at any time t is
d 2 r d fdr\ d
dt 2 ~ dt\dtj ~ eft (2« 61 + 12^)  2i + 24*k
At time t  1 this is 2i + 24k.
(b) The true acceleration at any time t is
At time t — 1 this equals
2i + 24k + (4i  2j + 12k) X (2i  6j + 12k)
+ (2i  2j + 2k) X (2i  6j + 4k)
+ (2i  j + 6k) X {(2i  j + 6k) X (2i  6j + 4k)}
= 2i + 24k + (48i  24j  20k) + (4i  4j  8k) + (141 + 212j + 40k)
= 40i + 184j + 36k
150 MOVING COORDINATE SYSTEMS [CHAP. 6
CORIOLIS AND CENTRIPETAL ACCELERATION
6.8. Referring to Problem 6.5, find (a) the Coriolis acceleration, (b) the centripetal
acceleration and (c) their magnitudes at time t = 1.
(a) From Problem 6.5 we have,
Coriolis acceleration = 2« X dr/dt = (4i  2j + 12k) X (2i  6 j + 12k)
= 48i  24j  20k
(b) From Problem 6.5 we have,
Centripetal acceleration = a. X (• X r) = (2i  j + 6k) X (32i + 4j  10k)
= 14i + 212j + 40k
(c) From parts (a) and (6) we have
Magnitude of Coriolis acceleration = V(48) 2 + (24)2 + (_ 2 o)2 = 4\/205
Magnitude of centripetal acceleration = V(14) 2 + (212)2 + ( 40 )2 = 2\/ll,685
MOTION OF A PARTICLE RELATIVE TO THE EARTH
6.9. (a) Express Newton's second law for the motion of a particle relative to an XYZ
coordinate system fixed in space (inertial system). (6) Use (a) to find an equation of
motion for the particle relative to an xyz system having the same origin as the XYZ
system but rotating with respect to it.
(a) If m is the mass of the particle (assumed constant), d 2 r/dt 2 \ F its acceleration in the fixed
system and F the resultant of all forces acting on the particle as viewed in the fixed system,
then Newton's second law states that
d 2 r
m
dt 2 \F
= F (1)
(b) Using subscript M to denote quantities as viewed in the moving system, we have from
Problem 6.6,
dt 2
d?r I .... . „ w dt
§§ +iXr+2.X^I +.X(.Xr) («)
dt 2 \m dt \m
Substituting this into (1), we find the required equation
m^§ = F  m(iXr)  2w(. xfjl )  m[. X (. X r)] (*)
dt 2 m \ dt \m j
We can drop the subscript M provided it is clear that all quantities except F are as
determined by an observer in the moving system. The quantity F, it must be emphasized, is the
resultant force as observed in the fixed or inertial system. If we do remove the subscript M
and write dr/dt = v, then (3) can be written
= F  w(i X r)  2m(» X v) — m[» X (*> X r)] (4)
d 2 v
6.10. Calculate the angular speed of the earth about its axis.
Since the earth makes one revolution [2tt radians] about its axis in approximately 24 hours =
86,400 sec, the angular speed is
w = 86^400 = 7.27 X 105 rad/sec
The actual time for one revolution is closer to 86,164 sec and the angular speed 7.29 X lO" 5 rad/sec.
CHAP. 6]
MOVING COORDINATE SYSTEMS
151
MOVING COORDINATE SYSTEMS IN GENERAL
6.11. Work Problem 6.4 if the origins of the XYZ
and xyz systems do not coincide.
Let R be the position vector of origin Q of the
xyz system relative to origin O of the fixed (or inertial)
XYZ system [see Fig. 63]. The velocity of the par
ticle P relative to the moving system is, as before,
dr\ _ dv
dt \m dt
dx. . dy . . dz .
i + "rrj + — k
dt
dt
dt
(1)
Now the position vector of P relative to O is
P = R + r and thus the velocity of P as viewed in the
XYZ system is
dp
dt ~
di (R + r) \ F " H\f + di
dv
= * + dt + » Xr ^
using equation (3) of Problem 6.4. Note that R is the
velocity of Q with respect to O. If R = this re
duces to the result of Problem 6.4.
Fig. 63
6.12. Work Problem 6.6 if the origins of the XYZ and xyz systems do not coincide.
Referring to Fig. 63, the acceleration of the particle P relative to the moving system is,
as before,
= dfr &x i+ d*yd?z
m dt 2 dt 2 ^ dt 2 ' dt 2 K1)
d?r
dt 2
Since the position vector of P relative to O is p = R + r, the acceleration of P as viewed in the
XYZ system is
d 2 p
dt 2
J^+'i
dm,
dt 2
d 2 r
F dt 2
= R + S + ^Xr + 2.X*+«X( tt Xr)
dt 2
dt
dt
(2)
using equation (S) of Problem 6.6. Note that R is the acceleration of Q with respect to O. If
R = this reduces to the result of Problem 6.6.
6.13. Work Problem 6.9 if the origins of the XYZ and xyz systems do not coincide.
(a) The position vector of the particle relative to the fixed (XYZ) system is p. Then the
required equation of motion is
d?P
dt 2
(1)
(6) Using the result (2) of Problem 6.12 in (1), we obtain
d 2 r
dt 2
= F
mR — m(i X r) — 2m(<* X v) — m[» X (o> X r)]
where F is the force acting on m as viewed in the inertial system and where v = r.
6.14. Find the equation of motion of a particle relative to an observer on the earth's
surface.
152
MOVING COORDINATE SYSTEMS
[CHAP. 6
We assume the earth to be a sphere with center
at [Fig. 64] rotating about the Z axis with
angular velocity » = wK. We also use the fact that
the effect of the earth's rotation around the sun is
negligible, so that the XYZ system can be taken as
an inertial system.
Then we can use equation (2) of Problem 6.12.
For the case of the earth, we have
m = (1)
R = o> X (o> X R) (2)
F = 2»E, ,„
the first equation arising from the fact that the ro
tation of the earth about its axis proceeds with con
stant angular velocity, the second arising from the
fact that the acceleration of origin Q relative to O
is the centripetal acceleration, and the third arising
from Newton's law of gravitation. Using these in
(2) of Problem 6.12 yields the required equation,
Fig. 64
dt*
GM
p — a) X (*» X R) — 2(<* X v)  « X (» X r)
(4)
assuming that other forces acting on m [such as air resistance, etc.] are neglected.
We can define
GM .
gp — tt X («* X R)
as the acceleration due to gravity, so that (4) becomes
^ = g  2(» X v)  » X (• X r)
(5)
(6)
Near the earth's surface the last term in (6) can be neglected, so that to a high degree of
approximation ,
= g  2(» X v)
(7)
In practice we choose g as constant in magnitude although it varies slightly over the earth's
surface. If other external forces act, we must add them to the right side of equations (6) or (7).
6.15. Show that if the particle of Problem 6.14 moves near the earth's surface, then the
equations of motion are given by
x = 2«> cos A y
y = — 2(w cos A x + m sin A z)
z = —g + 2o>sinXy
where the angle A is the colatitude [see Fig. 64] and 90°  A is the latitude.
From Fig. 64 we have
K = (Ki)i + (Kj)j + (Kk)k
= (—sin X)i + Oj + (cos X)k = sin X i + cos X k
an( j s0 «» = mK = —w sin X i + w cos X k
CHAP. 6] MOVING COORDINATE SYSTEMS 153
Then » X v = » X (xi + yj + zk)
i j k
— <o sin X w cos \
• • •
x y z
= (— w cos X y)\ + (u cos \ x + u sin X z)j — (« sin X y)k
Thus from equation (7) of Problem 6.14 we have
= — gk. + 2« cos X y i — 2(« cos \ x + a sin X z)j + 2<o sin X 2/ k
Equating corresponding coefficients of i, j,k on both sides of this equation, we find, as required,
x = 2w cos X y (1)
y = — 2(u cos X x + <o sin X z) (2)
z — — g + 2u sin X y (3)
6.16. An object of mass m initially at rest is dropped to the earth's surface from a height
which is small compared with the earth's radius. Assuming that the angular speed
of the earth about its axis is a constant «>, prove that after time t the object is
deflected east of the vertical by the amount %<»gt 3 sin A.
Method 1.
We assume that the object is located on the z axis at * = 0, y = 0, z — h [see Fig. 64]. From
equations (i) and (2) of Problem 6.15 we have on integrating,
x = 2(o cos X y + c it V — — 2(w cos X x + <o sin X z) + c 2
Since at t = 0, x = 0, y = 0, x = 0, y = 0, z = ft we have c x = 0, c 2 = 2u sin X ft. Thus
x = 2<o cos X j/, y = — 2(w cos X x + w sin X z) + 2a sin X ft (i)
Then (5) of Problem 6.15 becomes
'z = — g + 2w sin X y — —g — 4u 2 sin X [cos X a; + sin X (z — h)]
But since the terms on the right involving u 2 are very small compared with — g we can neglect them
and write z = —g. Integration yields z = —gt + c 3 . Since 1 = at t = 0, we have c 3 = or
z = gt (2)
Using equation (2) and the first equation of (1) in equation (2) of Problem 6.15 we find
j/ = (—2(0 cos X)(2w cos X 2/) + (— 2u sin X)(—gt)
= — 4w 2 cos 2 X 1/ + 2w sin X gt
Then neglecting the first term, we have y = 2« sin X gt. Integrating,
y — wg sin X t 2 + c 4
Since y = at £ = 0, we have c 4 = and y — ag sin X t 2 . Integrating again,
2/ = ^ug sin X t 3 + c 5
Then since y = at £ = 0, c 5 = so that, as required,
y  ^ug sin X t 3 (3)
Method 2.
Integrating equations (!), (£) and (£) of Problem 6.15, we have
x = 2w cos X 2/ + c x
y = — 2(w cos X x + (o sin X z) + c 2
z = —gt + 2w sin \y + c z
154
MOVING COORDINATE SYSTEMS
[CHAP. 6
Using the fact that at t = 0,
c 2 = 2o>h sin X, c 3 = 0. Thus
x = y — z = and
0; y = 0,
h, we have e x = 0,
2w cos X i/
— 2(<o cos X » + w sin X 2) + 2uh sin X
— gt + 2w sin X y
Integrating these we find, using the above conditions,
2w cos X I y du
J"
y — 2aht sin X — 2w cos XI x du — 2w sin X
r
2 dw
— h — igt 2 + 2w sin X
y du
(5)
(6)
Since the unknowns are under the integral sign, these equations are called integral equations.
We shall use a method called the method of successive approximations or method of iteration to
obtain a solution to any desired accuracy. The method consists of using a first guess for x, y, z
under the integral signs in (4), (5) and (6) to obtain a better guess. As a first guess we can try
x — 0, y = 0, z = under the integral signs. Then we find as a second guess
x = 0, 1/ = 2wfti sin X, z  h — ±gt 2
Substituting these in (4), (5) and (6) and neglecting terms involving <o 2 , we find the third guess
x  0, y = 2cofct sin X — 2<o sin X(7ii — ^g&)
— ^ug& sin X,
z — h
&*
Using these in (4), (5) and (6) and again neglecting terms involving w 2 , we find the fourth guess
0,
[ugt 3 sin X,
= h
i9t 2
Since this fourth guess is identical with the third guess, these results are accurate up to terms
involving w 2 , and no further guesses need be taken. It is thus seen that the deflection is
y = j^ugt 3 sin X, as required.
6.17. Referring to Problem 6.16, show that an object dropped from height h above the
earth's surface hits the earth at a point east of the vertical at a distance
%oih sin A yjlhlg.
From (2) of Problem 6.16 we have on integrating, z=_±gt 2 + c. Since zh at t  0, ch
and zh \g&. Then at z = 0, h = ±gt 2 or t = V^ft/fl'. Substituting this value of t into (5)
of Problem 6.16, we find the required distance.
THE FOUCAULT PENDULUM
6.18. Derive an equation of motion for a simple
pendulum, taking into account the earth's
rotation about its axis.
Choose the xyz coordinate system of Fig. 65.
Suppose that the origin O is the equilibrium position
of the bob B, A is the point of suspension and the
length of string AB is I. If the tension in the string
is T, then we have
T = (Ti)i + (Tj)j + (Tk)k
= T cos a i + T cos p j + T cos y k
' x\ . „, / y\ . , mil — z
= T
Ti l
j + T
I
(1)
Since the net force acting on B is T + mg, the equa
tion of motion of B is given by [see Problem 6.14]
m^J = T + mg  2m(« X v)  m<* X (• X r) (*)
Fig. 65
CHAP. 6] MOVING COORDINATE SYSTEMS 155
If we neglect the last term in (2), put g = — gk and use (1), then (2) can be written in component
form as
m x = —T(x/l) + 2muy cos \ (S)
my = —T(y/l) — 2mu(x cos \ + z sin X) (4)
m*z = T(l — z)/l — mg + 2mJy sin X (5)
6.19. By assuming that the bob of the simple pendulum in Problem 6.18 undergoes small
oscillations about the equilibrium position so that its motion can be assumed to take
place in a horizontal plane, simplify the equations of motion.
Making the assumption that the motion of the bob takes place in a horizontal plane amounts
to assuming that z and z are zero. For small vibrations (l — z)/l is very nearly equal to one.
Then equation (5) of Problem 6.18 yields
= T — mg + 2muy sin X
or T = mg — 2muy sin X (1)
Substituting (1) into equations (3) and (4) of Problem 6.18 and simplifying, we obtain
gx . 2axy sinX . „ . . ,„•>
x = — ^j H —j h 2wj/cosX (2)
gy , 2ayy sin X „ • . /ff v
y = —^j + — ^y 2 <oa; cosX (3)
These differential equations are nonlinear because of the presence of the terms involving xy
and yy. However, these terms are negligible compared with the others since «, x and y are
small. Upon neglecting them we obtain the linear differential equations
x = —gx/l + 2uy cos X (4)
y = —gy/l — 2(o* cos X (5)
6.20. Solve the equations of motion of the pendulum obtained in Problem 6.19, assuming
suitable initial conditions.
Suppose that initially the bob is in the yz plane and is given a displacement from the z axis
of magnitude A > 0, after which it is released. Then the initial conditions are
x = 0, x = 0, y = A, y = at t = (1)
To find the solution of equations (4) and (5) of Problem 6.19, it is convenient to place
K 2 = g/l, a = a cos X (2)
so that they become x — —K 2 x + 2ay (8)
y = K 2 y  2ax U)
It is also convenient to use complex numbers. Multiplying equation (4) by i and adding to (3),
we find
x + iy = K 2 (x + iy) + 2 a (yix) = K 2 (x + iy)  2ia(x + iy)
Then calling u = x + iy, this can be written
u = K 2 u  2iau or u + 2iau + K 2 u = (5)
If u = Cey* where C and y are constants, this becomes
y 2 + 2iay + K 2 =
so that V = (2i« ± V / 4a 2  4K 2 )/2 = ia ± Wa 2 + K 2 (6)
Now since a 2 = w 2 cos 2 X is small compared to K 2 = g/l, we can write
y = — ia ± iK (7)
156 MOVING COORDINATE SYSTEMS [CHAP. 6
Then solutions of the equation are (allowing for complex coefficients)
(C 1 + iC 2 )e Ka  K > t and (C 3 + iC 4 )e«« + K>t
and the general solution is
u = (C^iCJe*"™ +■ (C 3 + tC 4 )e«« + K)t (5)
where d> C 2 , C 3 , C 4 are assumed real. Using Euler's formulas
e i0 = costf + ism 6, e~ ie = cos e — t sin (9)
and the fact that u = x + iy, (8) can be written
x + iy = (C x + t'C 2 ){cos (a  K)t  i sin (a  K)£} + (C 3 + iC 4 ){cos (« + #)£  i sin (a + K)t)
Equating real and imaginary parts, we find
x = d cos («  #)* + C 2 sin («  K)t + C 3 cos (a + K)t + C 4 sin (a + X)t (10)
y = dam (a K)t + C 2 cos (a K)t  C 3 sin (a + K)t + C 4 cos (a + K)t (11)
Using the initial condition x = at * = 0, we find from (10) that d + C 3 = or
C 3 = d. Similarly, using x = at t = 0, we find from (10) that
X _ a \ /Vff/l wcosX
c * " C »U + J " C2 Vx^77+ w cosx,
Now since a cos X is small compared with \fgjl, we have, to a high degree of approximation,
C 4 = C 2 .
Thus equations (10) and (ii) become
x = d cos («  X)t + C 2 sin (a  K)t  d cos ( a + K ) 1 + C 2 sin (« + K)t (12)
y = d sin (a  K)t + C 2 cos (a  K)t + d sin (a + K)« + C 2 cos (a + X)t (15)
Using the initial condition y = 0, (1*) yields d = 0 Similarly using y = A at t = 0, we
find C 2 = A. Thus (1«) and (15) become
x = \A sin (a  K)t + A sin (a + K)t
y = \A cos (a  K)« + \A cos (a + #)*
or x = A cos Kt sm at "1
2/ = A COSK* COS at J
i e., a; = A cosy/gilt sin(ucosXt) 1
i (■*£)
j/ = A cos V^ t cos (w cos H) J
6.21. Give a physical interpretation to the solution (15) of Problem 6.20.
In vector form, (15) can be written
r = x\ + 2/j = A cos V 'gilt n
w here n = i sin (« cos X)t + j cos (w cos \)t
is a unit vector.
The period of cos yfgllt [namely, 2ttV^] is very small compared with the period of n [namely,
2tt/((ocosX)]. It follows that n is a very slowly turning vector. Thus physically the pendulum
oscillates in a plane through the z axis which is slowly rotating (or precessing) about the z axis.
Now at t = 0, n = j and the bob is at y = A. After a time t = 2^/(4 w cosX), for example,
n = l\/2i + JV^j so that the rotation of the plane is proceeding in the clockwise direction as
viewed from 2 above the earth's surface in the northern hemisphere [where cosX>0]. In the
southern hemisphere the rotation of the plane is counterclockwise.
The rotation of the plane was observed by Foucault in 1851 and served to provide laboratory
evidence of the rotation of the earth about its axis.
CHAP. 6]
MOVING COORDINATE SYSTEMS
157
MISCELLANEOUS PROBLEMS
6.22. The vertical rod AB of Fig. 66 is rotating with
constant angular velocity <■>. A light inextensible
string of length I has one end attached at point O
of the rod while the other end P of the string has a
mass m attached. Find (a) the tension in the string
and (b) the angle which string OP makes with the
vertical when equilibrium conditions prevail.
Choose unit vectors i and k perpendicular and parallel
respectively to the rod and rotating with it. The unit vector
j can be chosen perpendicular to the plane of i and k. Let
r = I sin e i — I cos e k
be the position vector of m with respect to O.
Three forces act on particle m
(i) The weight, mg = — mgk
(ii) The centrifugal force,
m{a X (» X r)} = m{[tok] X ([wk] X [I sin e i  I cos B k])}
= — m{[wk] X (al sin e j)} = ma 2 l sin i
(iii) The tension, T = —T sin e i + T cos 6 k
When the particle is in equilibrium, the resultant of all these forces is zero. Then
—mgk. + muH sin 6 i — T sin e i + T cos k =
i.e., (mu 2 l sin d — T sin $)i + (T cos 6 — mg)k =
or tow 2 / sin 6 — T sin  (1)
T cos 6 — mg = (2)
Solving (1) and (2) simultaneously, we find (a) T = mu 2 l, (6) o = cos 1 (g/u 2 l).
Since the string OP with mass m at P describe the surface of a cone the system is sometimes
called a conical pendulum.
Fig. 66
6.23. A rod AOB [Fig. 67] rotates in a vertical plane [the yz plane] about a horizontal
axis through O perpendicular to this plane [the x axis] with constant angular
velocity w. Assuming no frictional forces, determine the motion of a particle P of
mass m which is constrained to move along the rod. An equivalent problem exists
when the rod A OB is replaced by a thin hollow tube inside which the particle can move.
Fig. 67
At time t let r be the position vector of the particle and e the angle made by the rod with
the y axis. Choose unit vectors j and k in the y and z directions respectively and unit vector
i = j X k. Let rj be a unit vector in the direction r and e 1 a unit vector in the direction of
increasing e.
158 MOVING COORDINATE SYSTEMS [CHAP. 6
There are three forces acting on P:
(i) The weight, rag = — mgk — —mg sin r x — mg cos 9 X
(ii) The centrifugal force,
ra[o» X («* X r)] = ra[wi X («i X rr t )]
= — ra[coi(wi • rrj.) — rr 1 (wi • wi)]
= — ra[0 — wVrj] = rauVr!
(iii) The reaction force N = N8 t of the rod which is perpendicular to the rod since there are
no frictional or resistance forces.
Then by Newton's second law,
d 2 x
dt 2
d 2 x
ra^To — — ra^k + ra<o 2 rr x + 2V0J
or m7r^r 1 = —mg sine r t — mg cos 0j + mu 2 rr x + iVOj
= (ma 2 r — mg sin 0)r x + (N — mg cos ff)^
It follows that N = mg cos and
d 2 r/d« 2 = u 2 r  g sin (i)
Since 6 = a, a constant, we have 6 = wt if we assume = at t = 0. Then (i) becomes
d 2 r/d£ 2  w 2 r  sr sin u« (2)
If we assume that at t = 0, r = r , dr/dt = v , we find
or in terms of hyperbolic functions,
r = r cosh«t + ( — ^jsinhwt + ^sinw* (4)
6.24. (a) Show that under suitable conditions the particle of Problem 6.23 can oscillate
along the rod with simple harmonic motion and find these conditions, (b) What
happens to the particle if the conditions of (a) are not satisfied?
(a) The particle will oscillate with simple harmonic motion along the rod if and only if r =
and v = 0/2<o. In this case, r = (g/2<* 2 ) sin at. Thus the amplitude and period of the simple
harmonic motion in such case are given by g/2u> 2 and 2n/w respectively.
(6) If v  (g/2u) — ur then r = r e _wt + (g/2a 2 ) smut and the motion is approximately simple
harmonic after some time. Otherwise the mass will ultimately fly off the rod if it is finite.
6.25. A projectile located at colatitude X is fired with velocity v in a southward direction
at an angle « with the horizontal, {a) Find the position of the projectile after time t.
(b) Prove that after time t the projectile is deflected toward the east of the original
vertical plane of motion by the amount
^<*g sin Xt z — w^o cos (a — X) t 2
(a) We use the equations of Problem 6.15. Assuming the projectile starts at the origin, we have
x = 0, y = 0, z = at * = U)
Also, the initial velocity is v = v cos a i + v sin a k so that
x = v cos a, y = 0, z = v sin « at t = (2)
CHAP. 6] MOVING COORDINATE SYSTEMS 159
Integrating equations (1), (2) and (3) of Problem 6.15, we obtain on using conditions (2),
x = 2oj cos X y + v cos a (3)
y — — 2(w cos X x + w sin X z) (4)
z = —gt + 2w sin X y + v sin a (5)
Instead of attempting to solve these equations directly we shall use the method of iteration
or successive approximations as in Method 2 of Problem 6.16. Thus by integrating and using
conditions (1), we find
c
x — 2wcosX I y du + (t; cosa)( (6)
= — 2« cos X I x du — 2u sin X I z du
(7)
z = (v sin a)t — ±gt 2 + 2w sin X I y du (8)
Jo
As a first guess we use x = 0, y = 0, 2 = under the integral signs. Then (6), (7) and (8)
become, neglecting terms involving w 2 ,
x = (v cos a)t (9)
y = (10)
z = (v sina)£  \g& (11)
To obtain a better guess we now use (9), (10) and (II) under the integral signs in (6), (7) and
(8), thus arriving at
X = (V COS a)t (12)
y = —uv cos (a — \)t 2 + ^wgt 3 sin X (13)
z — (i) sin a)t — Iflrt 2 (14)
where we have again neglected terms involving w 2 . Further guesses again produce equations
(12), (13) and (14), so that these equations are accurate up to terms involving « 2 .
(6) From equation (13) we see that the projectile is deflected toward the east of the xz plane
by the amount ^ugt s sin X — wi; cos (a — X) t 2 . If v = this agrees with Problem 6.16.
6.26. Prove that when the projectile of Problem 6.25 returns to the horizontal, it will be at
the distance
o>vl sin 2 a
— ;r£ — (3 cos a cos A + sin a sin A)
to the west of that point where it would have landed assuming no axial rotation
of the earth.
The projectile will return to the horizontal when z = 0, i.e.,
(v sin a)t — ^gt 2 — or t — (2v sin a)/g
Using this value of t in equation (13) of Problem 6.25, we find the required result.
160 MOVING COORDINATE SYSTEMS [CHAP. 6
Supplementary Problems
ROTATING COORDINATE SYSTEMS. VELOCITY AND ACCELERATION
6.27. An xyz coordinate system moves with angular velocity «* = 2i — 3j + 5k relative to a fixed or
inertial XYZ coordinate system having the same origin. If a vector relative to the xyz system
is given as a function of time t by A = sin t i — cos t j + e t k, find (a) dA/dt relative to
the fixed system, (b) dA/dt relative to the moving system.
Ans. (a) (6 cos t  3e*)i + (6 sin t  2e*)j + (3 sin t  2 cos t  e')k
(b) cos t i + sin t j — e _t k
6.28. Find d 2 A/dt 2 for the vector A of Problem 6.27 relative to (a) the fixed system and (6) the moving
system.
Ans. (a) (6 cos t  45 sin t + Ue^i + (40 cos *  6 sin t  lle^j
+ (10 sin t  23 cos t + 16e~')k
(b) — sin t i + cos t j + e _t k
6.29. An xyz coordinate system is rotating with angular velocity o» = 5i — 4j — 10k relative to a fixed
XYZ coordinate system having the same origin. Find the velocity of a particle fixed in the xyz
system at the point (3, 1, —2) as seen by an observer fixed in the XYZ system.
Ans. 18i20j + 17k
6.30. Discuss the physical interpretation of replacing <* by *> in (a) Problem 6.4, page 148, and
(6) Problem 6.6, page 149.
6.31. Explain from a physical point of view why you would expect the result of Problem 6.3, page 148,
to be correct.
6.32. An xyz coordinate system rotates with angular velocity « = cos t i + sin t j + k with respect to a
fixed XYZ coordinate system having the same origin. If the position vector of a particle is given
by r = sin t i  cos t j + tk, find (a) the apparent velocity and (6) the true velocity at any
time t. Ans. (a) cos t i + sin t j + k (6) (t sin t + 2 cos *)i + (2 sin t  t cos t)j
6.33. Determine (a) the apparent acceleration and (6) the true acceleration of the particle of
Problem 6.32.
Ans. (a) sin ti + cos t j (6) (2t cost  3 sin t)i + (3 cos t + 2t sin «)j + (1  t)k
CORIOLIS AND CENTRIPETAL ACCELERATIONS AND FORCES
6.34. A ball is thrown horizontally in the northern hemisphere, (a) Would the path of the ball,
if the Coriolis force is taken into account, be to the right or to the left of the path when it is not
taken into account as viewed by the person throwing the ball? (6) What would be your answer
to (a) if the ball were thrown in the southern hemisphere? Ans. (a) to the right, (6) to the left
6.35. What would be your answer to Problem 6.34 if the ball were thrown at the north or south poles?
6.36. Explain why water running out of a vertical drain will swirl counterclockwise in the northern
hemisphere and clockwise in the southern hemisphere. What happens at the equator?
6.37. Prove that the centrifugal force acting on a particle of mass m on the earth's surface is a
vector (a) directed away from the earth and perpendicular to the angular velocity vector o> and
(6) of magnitude mu> 2 R sin X where X is the colatitude.
6.38. In Problem 6.37, where would the centrifugal force be (a) a maximum, (6) a minimum?
Ans. (a) at the equator, (6) at the north and south poles.
6.39. Find the centrifugal force acting on a train of mass 100,000 kg at (a) the equator (b) colatitude 30°.
Ans. (a) 35.0 kg wt, (6) 17.5 kg wt
6.40. (a) A river of width D flows northward with a speed v at colatitude X. Prove that the left bank
of the river will be higher than the right bank by an amount equal to
(2D u v cos\)(g 2 + 4« V cos^) 1 ' 2
where w is the angular speed of the earth about its axis.
(6) Prove that the result in part (a) is for all practical purposes equal to (2Duv cos \)/g.
6.41. If the river of Problem 6.40 is 2 km wide and flows at a speed of 5 km/hr at colatitude 45°,
how much higher will the left bank be than the right bank? Ans. 2.9 cm
CHAP. 6] MOVING COORDINATE SYSTEMS 161
6.42. An automobile rounds a curve whose radius of curvature is p. If the coefficient of friction is /j,
prove that the greatest speed with which it can travel so as not to slip on the road is Vwff
6.43. Determine whether the automobile of Problem 6.42 will slip if the speed is 60 mi/hr, n = .05 and
(a) p = 500 ft, (6) p = 50 ft. Discuss the results physically.
MOTION OF A PARTICLE RELATIVE TO THE EARTH
6.44. An object is dropped at the equator from a height of 400 meters. If air resistance is neglected,
how far will the point where it hits the earth's surface be from the point vertically below the
initial position? Ans. 17.6 cm toward the east
6.45. Work Problem 6.44 if the object is dropped (a) at colatitude 60° and (6) at the north pole.
Ans. (a) 15.2 cm toward the east
6.46. An object is thrown vertically upward at colatitude X with speed v . Prove that when it
returns it will be at a distance westward from its starting point equal to (±av\ sin X)/Sg 2 .
6.47. An object at the equator is thrown vertically upward with a speed of 60 mi/hr. How far from
its initial position will it land? Ans. .78 inches
6.48. With what speed must the object of Problem 6.47 be thrown in order that it return to a point
on the earth which is 20 ft from its original position? Ans. 406 mi/hr
6.49. An object is thrown downward with initial speed v . Prove that after time t the object is
deflected east of the vertical by the amount
wv sin X t 2 + ^ug sin X t 3
6.50. Prove that if the object of Problem 6.49 is thrown downward from height h above the earth's
surface, then it will hit the earth at a point east of the vertical at a distance
^JP^ (Vv 2 + 2gh  v )H^v 2 Q + 2gh + 2v )
6.51. Suppose that the mass m of a conical pendulum of length I moves in a horizontal circle of
radius a. Prove that (a) the speed is ayfg/^/l 2 — a 2 and (6) the tension in the string is
mgly/l 2 — a 2 .
6.52. If an object is dropped to the earth's surface prove that its path is a semicubical parabola.
THE FOUCAULT PENDULUM
6.53. Explain physically why the plane of oscillation of a Foucault pendulum should rotate clockwise
when viewed from above the earth's surface in the northern hemisphere but counterclockwise in
the southern hemisphere.
6.54. How long would it take the plane of oscillation of a Foucault pendulum to make one complete
revolution if the pendulum is located at (a) the north pole, (6) colatitude 45°, (c) colatitude 85°?
Ans. (a) 23.94 hr, (6) 33.86 hr, (c) 92.50 hr
6.55. Explain physically why a Foucault pendulum situated at the equator would not detect the
rotation of the earth about its axis. Is this physical result supported mathematically? Explain.
MOVING COORDINATE SYSTEMS IN GENERAL
6.56. An xyz coordinate system rotates about the z axis with angular velocity o> = cos t i + sin t j
relative to a fixed XYZ coordinate system where t is the time. The origin of the xyz system
has position vector R = ti — j + t 2 k with respect to the XYZ system. If the position vector of
a particle is given by r = (Zt + l)i — 2tj + 5k relative to the moving system, find the (a) apparent
velocity and (6) true velocity at any time.
6.57. Determine (a) the apparent acceleration and (6) the true acceleration of the particle in
Problem 6.56.
6.58. Work (a) Problem 6.5, page 148, and (b) Problem 6.7, page 149, if the position vector of the
xyz system relative to the origin of the fixed XYZ system is R = t 2 i — 2tj + 5k.
162
MOVING COORDINATE SYSTEMS
[CHAP. 6
MISCELLANEOUS PROBLEMS
6.59. Prove that due to the rotation o f the earth about its axis the app arent weight of an object of
mass m at colatitude X is m^(g  u 2 R sin 2 x) 2 + ( w 2# s j n x cos X )2 w here R is the* radius of
the earth.
6.60.
6.61.
Prove that the angle /? which the apparent vertical at colatitude X makes with the true vertical
o 2 R sin X cos X
is given by tan /? =
g — u 2 R sin 2 X '
Explain physically why the true vertical and apparent vertical would coincide at the equator
and also the north and south poles.
6.62. A stone is twirled in a vertical circle by a string of length 10 ft. Prove that it must have a
speed of at least 20 ft/sec at the bottom of its path in order to complete the circle.
6.63. A car C [Fig. 68] is to go completely around the vertical
circular loop of radius a without leaving the track.
Assuming the track is frictionless, determine the height
H at which it must start.
6.64. A particle of mass m is constrained to move on a friction
less vertical circle of radius a which rotates about a fixed
diameter with constant angular speed w. Prove that the
particle will make small oscillations about its equilibrium
position with a frequency given by 2vau/\/ a 2 w 4 — g 2 .
6.65. Discuss what happens in Problem 6.64 if « = yfg/a.
6.66. A hollow cylindrical tube AOB of length 2a [Fig. 69]
rotates with constant angular speed « about a vertical
axis through the center O. A particle is initially at rest
in the tube at a distance & from O. Assuming no frac
tional forces, find (a) the position and (6) the speed of
the particle at any time.
6.67. (a) How long will it take the particle of Problem 6.66 to
come out of the tube and (&) what will be its speed as
it leaves? Ans. (a) — In (a + y/a 2 — b 2 )
Fig. 68
CJ>
~B
Fig. 69
6.68. Find the force on the particle of Problem 6.66 at any position in the tube.
6.69. A mass, attached to a string which is suspended from a fixed point, moves in a horizontal circle
having center vertically below the fixed point with a speed of 20 revolutions per minute.
Find the distance of the center of the circle below the fixed point. Ans. 2.23 meters
6.70. A particle on a frictionless horizontal plane at colatitude X is given an initial speed v in a
northward direction. Prove that it describes a circle of radius v /(2w cos X) with period n7(<o cos X).
6.71. The pendulum bob of a conical pendulum describes a horizontal circle of radius a. If the length
of the pendulum is I, prove that the period is given by 4ir 2 y/l 2 — a 2 /g.
6.72. A particle constrained to move on a circular wire of radius a and coefficient n is given an initial
velocity v . Assuming no other forces act, how long will it take for the particle to come to rest?
6.73. (a) Prove that if the earth were to rotate at an angular speed given by y2g/R where R is its
radius and g the acceleration due to gravity, then the weight of a particle of mass m would be
the same at all latitudes. (6) What is the numerical value of this angular speed?
Ans. (b) 1.74 X 10~3 rad/sec
CHAP. 6]
MOVING COORDINATE SYSTEMS
163
6.74. A cylindrical tank containing water rotates about its axis with constant angular speed &>
so that no water spills out. Prove that the shape of the water surface is a paraboloid of revolution.
6.75. Work (a) Problem 6.16 and (6) Problem 6.17, accurate to terms involving w 2 .
6.76. Prove that due to the earth's rotation about its axis, winds in the northern hemisphere traveling
from a high pressure area to a low pressure area are rotated in a counterclockwise sense when
viewed above the earth's surface. What happens to winds in the southern hemisphere?
6.77. (a) Prove that in the northern hemisphere winds from the north,
east, south and west are deflected respectively toward the west,
north, east and south as indicated in Fig. 610. (b) Use this to
explain the origin of cyclones.
6.78. Find the condition on the angular speed so that a particle will
describe a horizontal circle inside of a frictionless vertical cone
of angle a.
6.79. Work Problem 6.78 for a hemisphere.
6.80. The period of a simple pendulum is given by P. Prove that its period when it is suspended from
the ceiling of a train moving with speed v around a circular track of radius p is given by
PVp~9/y/v* + p 2g2.
6.81. Work Problem 6.25 accurate to terms involving w 2 .
6.82. A thin hollow cylindrical tube OA inclined at angle a with the
horizontal rotates about the vertical with constant angular speed
w [see Fig. 611]. If a particle constrained to move in this tube
is initially at rest at a distance a from the intersection O of the
tube and the vertical axis of rotation, prove that its distance r
from O at any time t is r = a cosh (ut cos a).
6.83. Work Problem 6.82 if the rod has coefficient of friction /i.
6.84. Prove that the particle of Problem 6.82 is in stable equilibrium
between the distances from O given by
g sin a ( 1 + n tan a
w 2 \ tan a — fi
assuming tan a < l//x.
g sin a / l — jit tan a
w 2 V tan a + n
and
Fig. 611
6.85.
6.86.
A train having a maximum speed equal to v is to round a curve with radius of curvature p. Prove
that if there is to be no lateral thrust on the outer track, then this track should be at a
height above the inner track given by av\l yjv\ + P 2 g 2 where a is the distance between tracks.
A projectile is fired at colatitude X with velocity v directed toward the west and at angle a with
the horizontal. Prove that if terms involving « 2 are neglected, then the time taken to reach
the maximum height is
v sin a 2uv sin X sin a cos a
g ff 2
Compare with the case where w = 0, i.e. that the earth does not rotate about its axis.
6.87. In Problem 6.86, prove that the maximum height reached is
v 2 sin 2 a 2uv% sin X sin 2 a cos a
2g g*
Compare with the case where <o = 0.
164 MOVING COORDINATE SYSTEMS [CHAP. 6
6.88. Prove that the range of the projectile of Problem 6.86 is
Vq sin 2a uVq sin a sin X (8 sin 2 a — 6)
+
3^2
Thus show that if terms involving co 2 and higher are neglected, the range will be larger, smaller
or the same as the case where <o = 0, according as a > 60°, a < 60° or a — 60° respectively.
6.89. If a projectile is fired with initial velocity v x i + v 2 j + v 3 k from the origin of a coordinate system
fixed relative to the earth's surface at colatitude X, prove that its position at any later time t
will be given by
x — v x t + uv 2 t 2 cos X
y — v 2 t — at 2 ^! cos \ + v 3 sin X) + ^ugt 3 sin X
z — v 3 t — ±gt 2 + uv 2 t 2 sin X
neglecting terms involving co 2 .
6.90. Work Problem 6.89 so as to include terms involving co 2 but exclude terms involving co 3 .
6.91. An object of mass m initially at rest is dropped from height h to the earth's surface at colatitude X.
Assuming that air resistance proportional to the instantaneous speed of the object is taken into
account as well as the rotation of the earth about its axis, prove that after time t the object is
deflected east of the vertical by the amount
2t ° ^ 3 " X [(g ~ 2hp 2 )(l  ePt) + phteBt  pgt + \g$ 2 t 2 \
neglecting terms of order w 2 and higher.
6.92. Work Problem 6.91, obtaining accuracy up to and including terms of order u 2 .
6.93. A frictionless inclined plane of length I and angle a located at colatitude X is so situated that
a particle placed on it would slide under the influence of gravity from north to south. If the
particle starts from rest at the top, prove that it will reach the bottom in a time given by
y g sin a
2ul sin X cos a
3g
and that its speed at the bottom is
yj2gl sin a — f ul sin a cos a sin X
neglecting terms of order co 2 .
6.94. (a) Prove that by the time the particle of Problem 6.93 reaches the bottom it will have undergone
a deflection of magnitude
2Zco / 21 , ...
5 \ : COS (a + X)
o \ g sin a
to the east or west respectively according as cos (a + X) is greater than or less than zero.
(&) Discuss the case where cos (a + X) = 0. (c) Use the result of (a) to arrive at the result
of Problem 6.17.
6.95. Work Problems 6.93 and 6.94 if the inclined plane has coefficient of friction (i.
Chapter 7 SYSTEMS
of PARTICLES
DISCRETE AND CONTINUOUS SYSTEMS
Up to now we have dealt mainly with the motion of an object which could be considered
as a particle or point mass. In many practical cases the objects with which we are concerned
can more realistically be considered as collections or systems of particles. Such systems
are called discrete or continuous according as the particles can be considered as separated
from each other or not.
For many practical purposes a discrete system having a very large but finite number
of particles can be considered as a continuous system. Conversely a continuous system can
be considered as a discrete system consisting of a large but finite number of particles.
DENSITY
For continuous systems of particles occupying a region of space it is often convenient to
define a mass per unit volume which is called the volume density or briefly density.
Mathematically, if AM is the total mass of a volume At of particles, then the density can
be defined as ,. aM m
<t = hm — — vU
at^o At
The density is a function of position and can vary from point to point. When the density
is a constant, the system is said to be of uniform density or simply uniform.
When the continuous system of particles occupy a surface, we can similarly define a
surface density or mass per unit area. Similarly when the particles occupy a line [or curve]
we can define a mass per unit length or linear density.
RIGID AND ELASTIC BODIES
In practice, forces applied to systems of particles will change the distances between
individual particles. Such systems are often called deformable or elastic bodies. In some
cases, however, deformations may be so slight that they may for most practical purposes
be considered nonexistent. It is thus convenient to define a mathematical model in which
the distance between any two specified particles of a system remains the same regardless
of applied forces. Such a system is called a rigid body. The mechanics of rigid bodies is
considered in Chapters 9 and 10.
DEGREES OF FREEDOM
The number of coordinates required to specify the position of a system of one or more
particles is called the number of degrees of freedom of the system.
Example 1.
A particle moving freely in space requires 3 coordinates, e.g. (x, y, z), to specify its position. Thus
the number of degrees of freedom is 3.
165
166
SYSTEMS OF PARTICLES
[CHAP. 7
Example 2.
A system consisting of N particles moving freely in space requires 3N coordinates to specify its
position. Thus the number of degrees of freedom is 32V.
A rigid body which can move freely in space has 6 degrees of freedom, i.e. 6 coordinates
are required to specify the position. See Problem 7.2.
CENTER OP MASS
Let n, r 2 , . . . , r N be the position vectors of a system of N particles of masses
m u m 2 , . . .,m N respectively [see Fig. 71].
The center of mass or centroid of the system of particles is defined as that point C
having position vector
mxxx + m 2 r 2 + • • • + m N r N _
~ M
r =
1 N
(2)
m 1 + m 2 + • • • + m N
N
where M = ^ m v is the total mass of the system. We sometimes use 2 or simply J)
in place of 2 •
..«2
Fig. 71
Fig. 72
For continuous systems of particles occupying a region % of space in which the
volume density is o, the center of mass can be written
■rdi
r =
J or dr
where the integral is taken over the entire region % [see Fig. 72]. If we write
f = xi + yj + zk, r„ = x v i + y v j + z v k
then (3) can equivalently be written as
and
 _ S WyXy _ 2) m v y v
X ~ M > $ ~ M '
JaXdr I aydr
<V  *J CD
X =
%
„, _ *"K
M ' ° M
where the total mass is given by either
M = ^m v
= \ <rdr
z =
M
JaZdr
M
M
(5)
(6)
(7)
CHAP. 7] SYSTEMS OF PARTICLES 167
The integrals in (3), (5) or (7) can be single, double or triple integrals, depending on
which may be preferable.
In practice it is fairly simple to go from discrete to continuous systems by merely
replacing summations by integrations. Consequently we will present all theorems for
discrete systems.
CENTER OF GRAVITY
If a system of particles is in a uniform gravitational field, the center of mass is some
times called the center of gravity.
MOMENTUM OF A SYSTEM OF PARTICLES
If v, = dxjdt = r„ is the velocity of m v , the total momentum of the system is defined as
N N
p = ]£ ra„v„ = ^m v r v (8)
v=l v=l
We can show [see Problem 7.3] that
p = Mv = M^ = Mi (9)
where v = di/dt is the velocity of the center of mass.
This is expressed in the following
Theorem 7.1. The total momentum of a system of particles can be found by multiplying
the total mass M of the system by the velocity v of the center of mass.
MOTION OF THE CENTER OF MASS
Suppose that the internal forces between any two particles of the system obey Newton's
third law. Then if F is the resultant external force acting on the system, we have [see
Problem 7.4] , „ ,_
F = ft = M w> = M dt < I0)
This is expressed in
Theorem 7.2. The center of mass of a system of particles moves as if the total mass
and resultant external force were applied at this point.
CONSERVATION OF MOMENTUM
Putting F = in (10), we find that
N
p = ^ m v \ v = constant (11)
v=l
Thus we have
Theorem 7.3. If the resultant external force acting on a system of particles is zero,
then the total momentum remains constant, i.e. is conserved. In such case the center of mass
is either at rest or in motion with constant velocity.
This theorem is often called the principle of conservation of momentum. It is a generaliza
tion of Theorem 28, page 37.
168 SYSTEMS OF PARTICLES [CHAP. 7
ANGULAR MOMENTUM OF A SYSTEM OF PARTICLES
The quantity N
O = 2 m v {r„Xv v ) (12)
is called the total angular momentum [or moment of momentum] of the system of particles
about origin 0.
THE TOTAL EXTERNAL TORQUE ACTING ON A SYSTEM
If F„ is the external force acting on particle v, then r„ x F„ is called the moment of the
force F v or torque about 0. The sum
N
A = 2 r v X F„ (iS)
is called the £o£aZ external torque about the origin.
RELATION BETWEEN ANGULAR MOMENTUM AND
TOTAL EXTERNAL TORQUE
If we assume that the internal forces between any two particles are always directed
along the line joining the particles [i.e. they are central forces], then we can show as in
Problem 7.12 that
da
A = dt W
Thus we have
Theorem 7.4. The total external torque on a system of particles is equal to the time
rate of change of the angular momentum of the system, provided the internal forces
between particles are central forces.
CONSERVATION OF ANGULAR MOMENTUM
Putting A = in (U), we find that
N
O = ^ m v (r v x v„) = constant (15)
Thus we have
Theorem 7.5. If the resultant external torque acting on a system of particles is zero,
then the total angular momentum remains constant, i.e. is conserved.
This theorem is often called the principle of conservation of angular momentum. It is the
generalization of Theorem 2.9, page 37.
KINETIC ENERGY OF A SYSTEM OF PARTICLES
The total kinetic energy of a system of particles is defined as
\ N j N
T = « Jm,t)J = sXm,^ (16)
WORK
If Tv is the force (external and internal) acting on particle v, then the total work done
in moving the system of particles from one state [symbolized by 1] to another [symbolized
by 2] is
W12 = £ I yWr„ (17)
v=l *s \
CHAP. 7] SYSTEMS OF PARTICLES 169
As in the case of a single particle, we can prove the following
Theorem 7.6. The total work done in moving a system of particles from one state
where the kinetic energy is 2\ to another where the kinetic energy is T 2 , is
W12 = T 2 T, (is)
POTENTIAL ENERGY. CONSERVATION OF ENERGY
When all forces, external and internal, are conservative, we can define a total potential
energy V of the system. In such case we can prove the following
Theorem 7.7. If T and V are respectively the total kinetic energy and total potential
energy of a system of particles, then
T + V = constant (19)
This is the principle of conservation of energy for systems of particles.
MOTION RELATIVE TO THE CENTER OF MASS
It is often useful to describe the motion of a system of particles about [or relative to]
the center of mass. The following theorems are of fundamental importance. In all cases
primes denote quantities relative to the center of mass.
Theorem 7.8. The total linear momentum of a system of particles about the center
of mass is zero. In symbols,
N N
Y, m v< = ^m v r f v = (20)
v=X v=l
Theorem 7.9. The total angular momentum of a system of particles about any point
equals the angular momentum of the total mass assumed to be located at the center of mass
plus the angular momentum about the center of mass. In symbols,
O = r X Mv + 2 m v (r' v X v' v ) (21)
v=l
Theorem 7.10. The total kinetic energy of a system of particles about any point O
equals the kinetic energy of translation of the center of mass [assuming the total mass
located there] plus the kinetic energy of motion about the center of mass. In symbols,
T = ^Mv 2 + 5>X 2 (22)
Theorem 7.11. The total external torque about the center of mass equals the time rate
of change in angular momentum about the center of mass, i.e. equation (lb) holds not
only for inertial coordinate systems but also for coordinate systems moving with the
center of mass. In symbols,
da'
A = W M
If motion is described relative to points other than the center of mass, the results
in the above theorems become more complicated.
IMPULSE
If F is the total external force acting on a system of particles, then
170 SYSTEMS OF PARTICLES [CHAP. 7
f
•St.
¥dt (U)
h
is called the total linear impulse or briefly total impulse. As in the case of one particle,
we can prove
Theorem 7.12. The total linear impulse is equal to the change in linear momentum.
Similarly if A is the total external torque applied to a system of particles about origin O,
then t2
( A eft (25)
is called the total angular impulse. We can then prove
Theorem 7.13. The total angular impulse is equal to the change in angular momentum.
CONSTRAINTS. HOLONOMIC AND NONHOLONOMIC CONSTRAINTS
Often in practice the motion of a particle or system of particles is restricted in some
way. For example, in rigid bodies [considered in Chapters 9 and 10] the motion must be such
that the distance between any two particular particles of the rigid body is always the same.
As another example, the motion of particles may be restricted to curves or surfaces.
The limitations on the motion are often called constraints. If the constraint condition
can be expressed as an equation
<£(ri,r 2 , ...,r w , t) = (26)
connecting the position vectors of the particles and the time, then the constraint is called
holonomic. If it cannot be so expressed it is called nonholonomic.
VIRTUAL DISPLACEMENTS
Consider two possible configurations of a system of particles at a particular instant
which are consistent with the forces and constraints. To go from one configuration to
the other, we need only give the vth particle a displacement Sr* from the old to the new
position. We call Sr, a virtual displacement to distinguish it from a true displacement
[denoted by dr v ] which occurs in a time interval where forces and constraints could be
changing. The symbol 8 has the usual properties of the differential d; for example,
S(sin0) = cos0 80.
STATICS OF A SYSTEM OF PARTICLES.
PRINCIPLE OF VIRTUAL WORK
In order for a system of particles to be in equilibrium, the resultant force acting on each
particle must be zero, i.e. F„ = 0. It thus follows that F, • 8r v = where F v • 8r v is called
the virtual work. By adding these we then have
2Fv8r„ = (27)
v = l
If constraints are present, then we can write
F„ = F l v a) + F< c) (28)
where F< a) and F< c) are respectively the actual force and constraint force acting on the vth
particle. By assuming that the virtual work of the constraint forces is zero [which is true
for rigid bodies and for motion on curves and surfaces without friction], we arrive at
CHAP. 7] SYSTEMS OF PARTICLES 171
Theorem 7.14. A system of particles is in equilibrium if and only if the total virtual
work of the actual forces is zero, i.e. if
2 F< 0) • 8r„ = (29)
v=l
This is often called the principle of virtual work.
EQUILIBRIUM IN CONSERVATIVE FIELDS.
STABILITY OF EQUILIBRIUM
The results for equilibrium of a particle in a conservative force field [see page 38]
can be generalized to systems of particles. The following theorems summarize the basic
results.
Theorem 7.15. If V is the total potential of a system of particles depending on
coordinates q v q 2 , . . . , then the system will be in equilibrium if
W^ ' 8o 2 = °> •• («)
Since the virtual work done on the system is
" = £". + £".♦•••
(31) is equivalent to the principle of virtual work.
Theorem 7.16. A system of particles will be in stable equilibrium if the potential V
is a minimum.
In case V depends on only one coordinate, say q v sufficient conditions are
dV A dW
Other cases of equilibrium where the potential is not a minimum are called unstable.
D'ALEMBERT'S PRINCIPLE
Although Theorem 7.14 as stated applies to the statics of a system of particles, it can be
restated so as to give an analogous theorem for dynamics. To do this we note that according
to Newton's second law of motion,
F„ = p„ or F„  p„ = (so)
where p v is the momentum of the vth particle. The second equation amounts to saying
that a moving system of particles can be considered to be in equilibrium under a force
F„  p v , i.e. the actual force together with the added force p„ which is often called the
reversed effective force on particle v. By using the principle of virtual work we can then
arrive at
Theorem 7.17. A system of particles moves in such a way that the total virtual work
2 (F^ a)  p.) • 8r„ =
With this theorem, which is often called D'Alembert's principle, we can consider dynamics
as a special case of statics.
172 SYSTEMS OF PARTICLES [CHAP. 7
Solved Problems
DEGREES OF FREEDOM
7.1. Determine the number of degrees of freedom in each of the following cases:
(a) a particle moving on a given space curve; (b) five particles moving freely in a
plane; (c) five particles moving freely in space; (d) two particles connected by a
rigid rod moving freely in a plane.
(a) The curve can be described by the parametric equations x = x(s), y = y(s), z = «(s) where
s is the parameter. Then the position of a particle on the curve is determined by specifying
one coordinate, and hence there is one degree of freedom.
(b) Each particle requires two coordinates to specify its position in the plane. Thus 5 • 2 = 10
coordinates are needed to specify the positions of all 5 particles, i.e. the system has 10 degrees
of freedom.
(c) Since each particle requires three coordinates to specify its position, the system has 5*3 = 15
degrees of freedom.
(d) Method 1.
The coordinates of the two particles can be expressed by {x lf y{) and (x 2 ,y 2 ), i.e. a total
of 4 coordinates. However, since the distance between these points is a constant a [the length
of the rigid rod], we have (x 1 — x 2 ) 2 + (l/i — 2/2) 2 = ft2 so that one of the coordinates can be
expressed in terms of the others. Thus there are 4 — 1 = 3 degrees of freedom.
Method 2.
The motion is completely specified if we give the two coordinates of the center of mass
and the angle made by the rod with some specified direction. Thus there are 2 + 1 = 3 degrees
of freedom.
72. Find the number of degrees of freedom for a rigid body which (a) can move freely
in three dimensional space, (b) has one point fixed but can move in space about
this point.
(a) Method 1.
If 3 noncollinear points of a rigid body are fixed in space, then the rigid body is also fixed
in space. Let these points have coordinates (x lt y x , zj, {x 2 , y 2 , z 2 ), (x 3 , y 3 , z 3 ) respectively, a total
of 9. Since the body is rigid we must have the relations
(x x  x 2 ) 2 + (y x  y 2 ) 2 + (z t  z 2 ) 2 = constant, (x 2  x 5 ) 2 + (y 2  y 3 ) 2 + {z 2  z 3 ) 2 = constant,
(x 3  x x ) 2 + (y 3  y x ) 2 + (z 3  z t ) 2 = constant
hence 3 coordinates can be expressed in terms of the remaining 6. Thus 6 independent
coordinates are needed to describe the motion, i.e. there are 6 degrees of freedom.
Method 2.
To fix one point of the rigid body requires 3 coordinates. An axis through this point
is fixed if we specify 2 ratios of the direction cosines of this axis. A rotation about the axis
can then be described by 1 angular coordinate. The total number of coordinates required,
i.e. the number of degrees of freedom, is 3 + 2 + 1 = 6.
(6) The motion is completely specified if we know the coordinates of two points, say (x^y^Zj)
and (x 2 ,y 2 ,z 2 ), where the fixed point is taken at the origin of a coordinate system. But since
the body is rigid we must have
x\ + y\ + z\ = constant, x\ + yl + z\= constant, (x x  x 2 ) 2 + (y t  y 2 ) 2 + {z t  z 2 ) 2 = constant
from which 3 coordinates can be found in terms of the remaining 3. Thus there are 3 degrees
of freedom.
CENTER OF MASS AND MOMENTUM OF A SYSTEM OF PARTICLES
7.3. Prove Theorem 7.1, page 167: The total momentum of a system of particles can be
found by multiplying the total mass M of the system by the velocity v of the center
of mass.
CHAP. 7] SYSTEMS OF PARTICLES 173
2 rn v v v
The center of mass is by definition, f = ^ — .
Then the total momentum is p — 2 i^ v v„ — 2 »»,, r„ = M dr/dt = Mx.
7.4. Prove Theorem 7.2, page 167: The center of mass of a system of particles moves
as if the total mass and resultant external force were applied at this point.
Let F„ be the resultant external force acting on particle v while f„ x is the internal force on
particle v due to particle \. We shall assume that f„„ = 0, i.e. particle v does not exert any
force on itself.
By Newton's second law the total force on particle v is
„ dPy J2
F " + ? f * = ~dt = dP> (w » r "> W
where the second term on the left represents the resultant internal force on particle v due to all
other particles.
Summing over v in equation (1), we find
if. —
d&
2 F v + 2 2 t v \ = m i 2 m v r v I (2)
Now according to Newton's third law of action and reaction, f„ x = — f Xl> so that the double
summation on the left of {2) is zero. If we then write
F = 2 F„ and f = ^^ m v r v (3)
V 1Y1 V
d 2 r
(2) becomes F = Mr^ (4)
Since F is the total external force on all particles applied at the center of mass f, the required
result is proved.
7.5. A system of particles consists of a 3 gram mass located at (1, 0, 1), a 5 gram mass
at (2, 1, 3) and a 2 gram mass at (3, 1, 1). Find the coordinates of the center of mass.
The position vectors of the particles are given respectively by
r t = i  k, r 2 = 2i + j + 3k, r 3 = 3i  j + k
Then the center of mass is given by
f _ 3(i  k) + 5(2i + j + 3k) + 2(3i  j + k) _ 1 . , 3 . , 7
3 + 5 + 2
Thus the coordinates of the center of mass are (—J*. A. 1)
v 1 0' 1 o '
3 + 5 + 2 10 1 + 10 j + 5 k
7.6. Prove that if the total momentum of a system is constant, i.e. is conserved, then the
center of mass is either at rest or in motion with constant velocity.
The total momentum of the system is given by
d ^ .. d r2^r„] rff
Then if p is constant, so also is dr/dt, the velocity of the center of mass.
7.7. Explain why the ejection of gases at high velocity from the rear of a rocket will move
the rocket forward.
Since the gas particles move backward with high velocity and since the center of mass does
not move, the rocket must move forward. For applications involving rocket motion, see Chapter 8.
174
SYSTEMS OF PARTICLES
[CHAP. 7
Ar„ = Ax v Ay v &z v
7.8. Find the centroid of a solid region % as in Fig. 73.
Consider the volume element At v of the solid. The
mass of this volume element is
AM V = o v At v = a v Ax v Ay v Az v
where a v is the density [mass per unit volume] and
Ax v , Ay v , Az„ are the dimensions of the volume element.
Then the centroid is given approximately by
2 r„ AM V 2 r v a v Ar v 2 *„ a v Ax v Ay v Az v
2 AM V 2 <r v At v 2 °v ^ x v ^Vv ^ z v
where the summation is taken over all volume elements
of the solid.
Taking the limit as the number of volume elements becomes infinite in such a way that
Ar„ » or Ax v » 0, Ay v » 0, Az v » 0, we obtain for the centroid of the solid:
dM  radr JJJ radxdydz
_ %
Fig. 73
9.
dx dy dz
where the integration is to be performed Over %, as indicated.
Writing r = xi + yj + zk, f = xi + yj + zk, this can also be written in component form as
\\\ yadxdydz jJj zadxdydz
% = _
III x a dx dy dz
%
%
III adxdy dz
9
Jlf*
dx dy dz
sss
dx dy dz
7.9. Find the centroid of the region bounded by the plane x + y + z = a and the planes
x = 0, y = 0, z = 0.
The region, which is a tetrahedron, is indicated in Fig. 74. To find the centroid, we use
the results of Problem 7.8.
In forming the sum over all volume elements of the region, it is advisable to proceed in an
orderly fashion. One possibility is to add first all terms corresponding to volume elements
contained in a column such as PQ in the figure. This amounts to keeping x v and y v fixed and
adding over all z v . Next keep *„ fixed but sum over all y v . This amounts to adding all columns,
such as PQ, contained in a slab RS, and consequently amounts to summing over all cubes contained
in such a slab. Finally, vary x v . This amounts to addition of all slabs such as RS.
In performing the integration over %, we use these
same ideas. Thus keeping x and y constant, integrate
from z = [base of column PQ] to z — a — x — y [top
of column PQ]. Next keep x constant and integrate
with respect to y. This amounts to addition of columns
having bases in the xy plane [z = 0] located anywhere
from R [where y = 0] to S [where x + y = a or y =
a — x], and the integration is from y = to y = a — x.
Finally, we add all slabs parallel to the yz plane, which
amounts to integration from x — to x — a. We
thus obtain
At„ = Ax v Ay v Az v
t =
Jr>a pa — x pa—x — y
I I a(xi + yj + zk) dz dy dx
x = *A/ = *z =
a — x /*a — x—i
na pa — x s*
K 'x=Q *^j/ = *^=0
a dz dy dx
Fig. 74
Since a is constant in this case, it may be cancelled. The denominator without a is evaluated
to be a 3 /6, and the numerator without a is (a 4 /24)(i + j + k). Thus the center of mass is
f = (a/4)(i + j + k) or x = a/4, y = a/4, z = a/4.
CHAP. 7]
SYSTEMS OF PARTICLES
175
7.10. Find the centroid of a semicircular region of radius a.
Method 1. Using rectangular coordinates.
Choose the region as in Fig. 75. The equation of the circle C is x 2 + y 2 = a 2 or y = y/a 2  x 2
since y ^ 0.
If a is the mass per unit area, assumed constant, then the coordinates of the centroid
are given by
3 =
J xadA JJ xdydx j j
y/l^x*
xdy dx
V =
f <rdA ff dydx C C
Jtf^
=
dydx
ydydx
/.
x = —a 3/=0
ydydx
odA
ss
dy dx
J._ J
V^?
dydx
2a 3 /3
ira 2 /2
4a
3lT
a "y=0
Note that we can write x  immediately, since by symmetry the centroid is on the y axis.
The denominator for y can be evaluated without integrating by noting that it represents the
semicircular area which is %ira 2 .
Fig. 75
Fig. 76
Method 2. Using polar coordinates.
The equation of the circle is r = a [see Fig. 76]. As before, we see by symmetry that the
centroid must lie on the y axis, so that x = 0. Since y = r sin e and dA=rdrde in polar
coordinates, we can write
y =
yadA I \ (r sin 9 ) r dr de
s
dA
%
r=0
rdrde
2q3/3
Aa
Sir
7.11. Find the center of mass of a uniform solid hemisphere of radius a.
By symmetry the center of mass lies on the
z axis [see Fig. 77]. Subdivide the hemisphere
into solid circular plates of radius r, such as
ABCDEA. If the center G of such a ring is at
distance z from the center O of the hemisphere,
r 2 + z 2 = a 2 . Then if dz is the thickness of the
plate, the volume of each ring is
irr 2 dz = ir{a 2 z 2 )dz
and the mass is ira(a 2 — z 2 ) dz. Thus we have
£
traz{a 2 — z 2 ) dz
2 — z =
J a
va(a 2 — z 2 ) dz
= la
z=0
Fig. 77
176 SYSTEMS OF PARTICLES [CHAP. 7
ANGULAR MOMENTUM AND TORQUE
7.12. Prove Theorem 7.4, page 168: The total external torque on a system of particles is
equal to the time rate of change of angular momentum of the system, provided that
the internal forces between particles are central forces.
As in equation (1) of Problem 7.4, we have
d P v d
x
Multiplying both sides of (1 ) by r v X , we have
f„ + 2f„x = of = Tt {m " Vv) {1)
r,XF v + 2 r„ X f„ x = v v X ^(m„ v„) (*)
d d
Since r„ X ^(m„v„) = ^K(r,Xv v )} (3)
(2) becomes r,XF, + 2 *„ X f vX = ^K(r,Xv,)} (4)
Summing over j» in (4), we find
2r,XF, + 2 2r„Xf vX = j i 2 m v (t v X v„) f (5)
l> l> X CtC ^ j; J
Now the double sum in (5) is composed of terms such as
r v XU + *xXf x „ («)
which becomes on writing £ X/ = f„ x according to Newton's third law,
r„Xf„ x  r x Xf„ x = (r„  r x ) X f vX (7)
Then since we suppose that the forces are central, i.e. f„ x has the same direction as r„ — r x , it follows
that (7) is zero and also that the double sum in (5) is zero. Thus equation (5) becomes
2r v XF„ = i\^m v (r v Xv v )\ or A = d °
dt \f "" v ^ v " *" ; J " dt
where A = 2 *i/ X F„, O = 2 Wj/(r„ X v„).
WORK, KINETIC ENERGY AND POTENTIAL ENERGY
7.13. Prove Theorem 7.6, page 169: The total work done in moving a system of particles
from one state to another with kinetic energies 2\ and T 2 respectively is T2T1.
The equation of motion of the pth particle in the system is
(1)
(2)
T v
= F„ + 2U =
d / • X
^(m,r,)
Taking the dot product of both sides with r„, we have
Tv'K ==
F„ • r v + 2 t v \ ' K z
d i • x
Since r„ • ^ (ra„ r„)
Id, ,• • >.
Id/ 2\
= 2 5t M
(2) can be written
Tv'K =
F„ • r„ + 2 f i>x * K
X
Id/ 2\
= 2di {m ^
(3)
2^r, = 2F„.r„ + 22Ur, = ^(2™,^) U)
Summing over y in equation (5), we find
CHAP. 7] SYSTEMS OF PARTICLES 177
Integrating both sides of (4) with respect to t from t = t x to t = t 2 , we find
w 12 = 2 ('r v 'r v dt = 2 ( 2 F v l v dt + 22 ( 2 t vX r v dt
ti »i l l
Using the fact that r„ <Zt = dr„ and the symbols 1 and 2 for the states at times t x and t 2
respectively, this can be written
W 12 = 2 Cr v 'dr v = 2 ( 2 F v 'dr v + 22 f f„x ' *„ = T 2  T x (5)
where Ti and T^ are the total kinetic energies at t x and t 2 respectively. Since
W
= 2 f f v 'dr v (6)
12
is the total work done (by external and internal forces) in moving the system from one state to
another, the required result follows.
It should be noted that the double sum in (5) indicating work done by the internal forces,
cannot be reduced to zero either by using Newton's third law or the assumption of central forces.
This is in contradistinction to the double sums in Problems 7.4 and 7.12 which can be reduced to zero.
7.14. Suppose that the internal forces of a system of particles are conservative and are
derived from a potential
Vxv (rw) = V v \ (r v \)
where n„ = r v \ = V( x * ~ x ») 2 + (#*. — Vv) 2 + (z\  z v ) 2 is the distance between par
ticles A and v of the system.
(a) Prove that J Jf^dr, = « 2 2 dVw where f x „ is the internal force on
v X £t v \
particle v due to particle A.
(b) Evaluate the double sum 22 f f\v'dr v of Problem 7.13.
(a) The force acting on particle v is
Wxv. Wxv. dV Xv
The force acting on particle X is
BV Xv ^ dV Xv dV Xv
hv  ~~B^ 1 ~ ~d^ j ~ ~J^ k = ~S^d x V Xv = V x V Xv = i vX (2)
The work done by these forces in producing the displacements dr v and dr x of particles v and X
respectively is
i vX 'dr v + i Xv 'dr x =
JdV Xv d v Kv dV Xv dV Xv dV Xv dV Xv , ]
= ~dV Xv
Then the total work done by the internal forces is
22fx„rfr l , = §22<n\, (s)
v X * v \
the factor £ on the right being introduced because otherwise the terms in the summation
would enter twice.
178
SYSTEMS OF PARTICLES
[CHAP. 7
(6) By integrating (3) of part (a), we have
2
2 2) t Xy • dr v =   2 2 I
C*Vx
where y< lnt) and y 2 mt) denote the total internal potentials
J22v x
at times ^ and £ 2 respectively.
2 t
1/(int) r/(int)
K l K 2
U)
(5)
7.15. Prove that if both the external and internal forces for a system of particles are
conservative, then the principle of conservation of energy is valid.
If the external forces are conservative, then we have
F„  VV„
(1)
from which
2 (\dr v  2 fV, = v?
T/(ext)
K 2
where V[ ext) and y< ext) denote the total external potential
2V„
V
at times t t and £ 2 respectively.
Using {2) and equation (4) of Problem 7.14(6) in equation (5) of Problem 7.13, we find
t 2  t x = y{ ext)  v< ext) + v< Int)  y< int) = v x  y 2 (*)
where
Vi
V
(ext)
+ vj
(int)
and
Vo
y^ ext) + y 2
(int)
U)
are the respective total potential energies [external and internal] at times t x and t % . We thus find
from (5),
T t + V x  T 2 + y 2 or T + V = constant (5)
which is the principle of conservation of energy.
MOTION RELATIVE TO THE CENTER OF MASS
7.16. Let r^ and \' v be respectively the position vector and velocity of particle v relative to
the center of mass. Prove that (a) ^ m v r' v = 0, (b) ^ m v \' v = 0.
V v
(a) Let r^ be the position vector of particle v relative to O
and f the position vector of the center of mass C relative
to O. Then from the definition of the center of mass,
r = ^^m v r v (1)
where M — 2 m v From Fig. 78 we have
r v = r ' v + f (2) Fig. 78
Then substituting (2) into (1), we find
from which 2 m v r v ~ ®
v
(b) Differentiating both sides of (3) with respect to t, we have 2 m* x v = °
(S)
CHAP. 7] SYSTEMS OF PARTICLES 179
7.17. Prove Theorem 7.9, page 169: The total angular momentum of a system of particles
about any point O equals the angular momentum of the total mass assumed to be
located at the center of mass plus the angular momentum about the center of mass.
Let r„ be the position vector of particle v relative to O, f the position vector of the center of
mass C relative to O and r' the position vector of particle v relative to C. Then
*v = <+* tt)
Differentiating with respect to t, we find
v„ = r„ = i' v + r = \' v + v (2)
where v is the velocity of the center of mass relative to O, v„ is the velocity of particle v relative to
O, and v' v is the velocity of particle v relative to C.
The total angular momentum of the system about is
a = Sw.fr.Xv,,) = 2^{« + f)X( V ; + v)}
V V
= 2 m v (r' v Xvp + 2 m,(r' X v) + 2 m v (t X V v ) + 2 m v {l X v) (3)
V V V V
Now by Problem 7.16,
K(<Xv) = J2^v<fXv =
2 m v {f Xv) = f X j 2 rn p \'V =
2 m v (t X v) = < 2 «H, Mr X v) = M{i X v)
Then (3) becomes, as required,
O = 2m„(<Xv') + Af(fXv)
7.18. Prove Theorem 7.10, page 169: The total kinetic energy of a system of particles
about any point O equals the kinetic energy of the center of mass [assuming the
total mass located there] plus the kinetic energy of motion about the center of mass.
The kinetic energy relative to O [see Fig. 78] is
* v *• V
Using equation (2) of Problem 7.16 we find
r v = r + r„ = v + v„
Thus (1) can be written
T = ^2^{(v + v;).(v + V ;)}
6 v
= «2^vv + 2™„v'< + h^^K'K
*• v v & v v
= (?^)v + v.jsm.v;} + §2*^;
= gMv 2 +  2 m„v' v
since 2 ™> v v ' v ~ by Problem 7.16.
,2
180 SYSTEMS OF PARTICLES [CHAP. 7
IMPULSE
7.19. Prove Theorem 7.12: The total linear impulse is equal to the change in linear
momentum.
The total external force by equation (4) of Problem 7.4 is
_, ifdH ,.dv
F = M di? = M lt
Then the total linear impulse is
X* 2 C H dv _
Ydt = I M ~^ dt ~ M *2  M\ 1 = p 2  Pi
where p x = Mx t and p 2 — Mv 2 represent the total momenta at times t x and t 2 respectively.
CONSTRAINTS. HOLONOMIC AND NONHOLONOMIC CONSTRAINTS
7.20. In each of the following cases state whether the constraint is holonomic or non
holonomic and give a reason for your answer: (a) a bead moving on a circular wire;
(b) a particle sliding down an inclined plane under the influence of gravity; (c) a
particle sliding down a sphere from a point near the top under the influence of gravity.
(a) The constraint is holonomic since the bead, which can be considered a particle, is constrained
to move on the circular wire.
(6) The constraint is holonomic since the particle is constrained to move along a surface which is
in this case a plane.
(c) The constraint is nonholonomic since the particle after reaching a certain location on the
sphere will leave the sphere.
Another way of seeing this is to note that if r is the position vector of the particle
relative to the center of the sphere as origin and a is the radius of the sphere, then the
particle moves so that r 2 § a 2 . This is a nonholonomic constraint since it is not of the
form (26), page 170. An example of a holonomic constraint would be r 2 = a 2 .
STATICS. PRINCIPLE OF VIRTUAL WORK. STABILITY
7.21. Prove the principle of virtual work, Theorem 7.14, page 171.
For equilibrium, the net resultant force F v on each particle must be zero, so that
2F„'8r„ = (0
V
But since F„ = F (a) + F (c) where F^ a) and F< c) are the actual and constraint forces acting on the
pth particle, (1) can be written
2F£ a) 8r„ + 2Ft c), 8'v = ° (2)
If we assume that the virtual work of the constraint forces is zero, the second sum on the left of
(2) is zero, so that we have _ , ^ / a \
2 F Co) • dr v = W
v v
which is the principle of virtual work.
7.22. Two particles of masses mi and m 2 are located on a frictionless double incline and
connected by an inextensible massless string passing over a smooth peg [see Fig. 79
below]. Use the principle of virtual work to show that for equilibrium we must have
sin a x _ m2
sin a 2 mi
where a x and a 2 are the angles of the incline.
CHAP. 7]
SYSTEMS OF PARTICLES
181
Method 1.
Let r x and r 2 be the respective position vectors
of masses m 1 and m 2 relative to O.
The actual forces (due to gravity) acting on
m 1 and m 2 are respectively
F< a) = m xS ,
F< a) = m 2 g
According to the principle of virtual work,
2Fj 0) «r v 
or F'< 0) • Sr x + F 2 a) • Sr 2 =
00
(*)
A,
m£
^\B
S\ a i
Wlj
Fig. 79
where 5r x and Sr 2 are the virtual displacements of m 1 and m 2 down the incline. Using (1) in (2),
™i£ • Sr x + wi 2 g • Sr 2 = (3)
or m x g 8r t sin a x + m 2 g 8r 2 sin a 2 = (4)
Then since the string is inextensible, i.e. 8r t + 8r 2 = or 8r 2 = —Sri, (4) becomes
(mrf sin a x — m^ sin a 2 )S^i =
But since 8r x is arbitrary, we must have m x g sin « x — m 2 g sin a 2 = 0> ie»>
sin ax _ m 2
h
(5)
sm a 2 w
Method 2.
When it is not clear which forces are constraint forces doing no work, we can take into account
all forces and then apply the principle of virtual work. Thus, for example, taking into account
the reaction forces R x and R 2 due to the inclines on the particles and the tension forces T x and T 2 ,
the principle of virtual work becomes
(m x g + T x + Rj) • 8r x + (m 2 g + T 2 + R 2 ) • Sr 2 =
Now since the inclines are assumed smooth [so that
the reaction forces are perpendicular to the in
clines] we have
Ri'Srj = 0, R 2 *8r 2 = (7)
Also, since there is no friction at the peg, the ten
sions Tj and T 2 have the same magnitude. Thus
we have, using the fact that 8r x and Sr 2 are directed
down the corresponding inclines and the fact that
8r 2 = 8r lt
T a • 8r x + T 2 • Sr 2 = T x 8r x  T 2 8r 2
= {T 2 T x )8r x = (8)
since T x = T 2 . Then using (7) and (5), (6) becomes
m x g • 5r x + m 2 g • Sr 2 =
as obtained in (3). Fig. 710
<*)
7.23. Use Theorem 7.15, page 171, to solve Problem 7.22.
Let the string have length I and suppose that the lengths of string OA and OB on the inclines
[Fig. 79] are x and l — x respectively. The total potential energy using a horizontal plane
through O as reference level is
V =
Then for equilibrium we must have
m x gx sin a t — m 2 g(l — x) sin a 2
— = —m x g sin a x + m^g sm a 2 =
sinaj
sin a 2
m 2
It should be noted that V is not a minimum in this case so that the equilibrium is not stable,
as is also evident physically.
182 SYSTEMS OF PARTICLES [CHAP. 7
D'ALEMBERTS PRINCIPLE
7.24. Use D'Alembert's principle to describe the motion of the masses in Problem 7.22.
We introduce the reversed effective forces m x r x and m 2 V 2 in equation (3) of Problem 7.22 to
obtain .. ..
(m x g m x r x )'8r x + (m 2 g  ra 2 r 2 ) • Sr a = (1)
This can be written #>
(TOjfli sin**! — m x r x )8r x + (ra 2 sina 2 — m 2 'r 2 )Sr 2 = (2)
Now since the string is inextensible so that r x + r % — constant, we have
8r x + 8r 2 = 0, r x + V 2 =
or Sr 2 = — Srj, r 2 = — rV Thus (2) becomes, after dividing by 8r x # 0,
wijflf sinaj — m^ — m 2 flr sina 2 — m 2 r x =
TOjjf sin a x — m^ sin a 2
n =
m x + m 2
Thus particle 1 goes down or up the incline with constant acceleration according as
m x g sin a x > m<$ sin a 2 or m x g sin a x < m^g sin a 2 respectively. Particle 2 in these cases goes
up or down respectively with the same constant acceleration.
We can also use a method analogous to the second method of Problem 7.22.
MISCELLANEOUS PROBLEMS
7.25. Two particles having masses mi and m 2 move
so that their relative velocity is v and the veloc
ity of their center of mass is v. If M = m\ + w 2
is the total mass and ju, = miw 2 /(wi + ra 2 ) is the
reduced mass of the system, prove that the total
kinetic energy is $Mv 2 + $ixv 2 .
Let r x , r 2 and f be the position vectors with re
spect to O of mass m x , mass m 2 and the center of
mass C respectively.
From the definition of the center of mass, we have
m 1 T 1 + m 2 r 2
t = and * — i
r m x + m 2 ™i + ™2
or using v x = i x , v 2 = r 2 , v = f , _
m x \ x + m 2 \ 2 = (m x + m 2 )v K 1 )
If the velocity of m x relative to m 2 is v, then
v = ^(r!r 2 ) = h  f 2 = v x  v 2
so that v i  v 2 = v
Solving (1) and (2) simultaneously, we find
m 2 v _ _ _ m i v
v i = V + «, + «,' v 2  v Wi + m2
Then the total kinetic energy is
T = l^ivf + 2 m 2V 2
m 2 v \ 2 l /_ m i v
^1% ... _ 1 a^o , 1
1 / w 2 v \ , 1 /_ '"''
(*)
= !<+«*■ +1^ = i w + i
/iV 2
CHAP. 7]
SYSTEMS OF PARTICLES
183
7.26. Find the centroid of a uniform semicircular
wire of radius a.
By symmetry [see Fig. 712] the centroid of
the wire must be on the y axis, so that x = 0. If o
is the mass per unit length of the wire, then if ds
represents an element of arc, we have ds = ado
so that
I y a ds i
J ads
r
(a sin e)(a de)
c
2a?
■na
f
de
2a
Fig. 712
7.27. Suppose that n systems of particles be given having centroids at f i, f 2 , . . . , f» and
total masses M x , M 2 , ... .,M n respectively. Prove that the centroid of all the systems
is a t
Mif l + M 2 i 2 + • • • + M n r n
Mi + M 2 +
+ M n
Let system 1 be composed of masses m n , m 12 , . . . located at r n , r 12 , . . . respectively. Similarly
let system 2 be composed of masses m 21 , m 22 , . . . located at r 21 , r 22 , .... Then by definition,
m n r n
+
m vFn
+ •••
m n
+
m 12 +
™21*21
+
w 22 r 22
+ •••
m 21
+
m 22 +
m nl r nl
+
m n2 r n2
+ '••
*2
™„i + in n2 +
But the centroid for all systems is located at
(m n r u + m 12 r 12 +•••) + (m 21 r 21 + m 22 r 22 +
ra u r u
+
^12*12
+ •••
Mi
m 21 r 21
+
w 22 r 22
+ •••
M 2
^nl^nl
+
m n2fn2
+ •••
M m
■ ) + • • • + (m nl r Ml + m n2 r n2 + • • • )
(w u + m 12 + • • ■) + (w 21 + m 22 + • • •) + • • • + (m nl + m n2 H )
Mjt x + M 2 r 2 + • • • + M n r n
M x + M 2 + • • • + Af „
7.28. Find the centroid of a solid of constant density
consisting of a cylinder of radius a and height H
surmounted by a hemisphere of radius a [see
Fig. 713].
Let f be the distance of the centroid of the solid from the
base. The centroid of the hemisphere of radius a is at dis
tance fa + H from the base of the solid, and its mass is
M x = ^a 3 £r [see Problem 7.11].
The centroid of the cylinder of radius a and height H
is at distance ±H from the base of the solid and its mass
is M 2 = ira 2 Ho.
Then by Problem 7.27,
(,ra3q)(fa + ff) + {TraWo){%H)
~ ira 3 a + 7ro 2 Hff ~
 3«2 + 8aH + 6m
~ 8a + 12H
Af,
/ ' T ■ "V
»
*r
>s>s> *~. ' ''iliiip'
4 /"»
r
1 ■#
i
•1
2 I ! «
! M 2
^ Base of s
olid
Fig. 713
184
SYSTEMS OF PARTICLES
[CHAP. 7
7.29. A circular hole of radius a/2 is cut out of a circular region of radius a, as shown
in Fig. 714. Find the centroid of the shaded region thus obtained.
y v
'x
y —%ira 2 o
x a
Fig. 714
Fig. 715
By symmetry the centroid is located on the x axis, so that y = 0.
We can replace the circular region of radius a by the mass M 1 — TraPo concentrated at its
centroid x x = a [Fig. 715]. Similarly, we can replace the circular hole of radius a/2 by the
negative mass M 2 = —\ira % o concentrated at its centroid x 2 = fa. Then the centroid is located
on the x axis at
M l x l + M 2 x 2 (Tra 2 a)(a) + (%ira 2 a)(%a) _ 5
Mi + M 2
ira 2 a — \ira % o
7.30. A uniform rod PQ [see Fig. 716] of mass m and length L has its end P resting
against a smooth vertical wall AB while its other end Q is attached by means of an
inextensible string OQ of length I to the fixed point on the wall. Assuming that
the plane of P, Q and O is vertical and perpendicular to the wall, show that
equilibrium occurs if
V4L 2  I 2
sin a
lV&
sm/3
There is only one actual force, i.e. the weight rag of the
rod. Other forces acting are the force of the wall on the
rod and the tension in the string. However, these are con
straint forces and can do no work. This can be seen since
if P were to slide down the wall no work would be done,
because the wall is frictionless and thus the force due to
the wall on the rod is perpendicular to the wall. Also if Q
were to drop, it could only move perpendicular to the
string at Q.
Let r be the position vector of the center of mass C
[in this case also the center of gravity] relative to 0. Also
let i and j be unit vectors in the horizontal and vertical
directions respectively so that r = xi + yj.
From Fig. 716,
OQ = OP + PQ
OQ = OC + CQ
V4L 2  I 2
Then from (1), on taking the dot product with i,
OQ«i = OP'i + PQ'i
Since OP • i = 0, this reduces to
OQ'i = PQ'i
or I sin a = L sin /?
Fig. 716
(1)
(2)
CHAP. 7]
SYSTEMS OF PARTICLES
185
Similarly on taking the dot product of both sides of (2) with j,
OQ'j = OC'j + CQ'j
or I cos a — y + \L cos /? (4)
Now a virtual displacement of the center of mass C is given by
Sr = 8xi + 8yj (5)
Since mg is the only actual force, the principle of virtual work becomes
mg • Sr = (6)
Using (5), this becomes mg 8y = or 8y = (7)
Now from (S) and (A), we have
I COS a 8a = L cos (i SfS
— I sin a 8a = 8y — \L sin S/3
since I and L are constants and since 8 has the same properties as the differential operator d.
Since 8y = from (7), these equations become
ZcosaSa = Lcos/JS/3 (5)
J sin a So; = lLsin/?S/? (9)
From (5) and (P), we have on division,
Now from (3),
so that
Thus equation (10) can be written
sin « _ 1 sin/?
cos a ~ 2 cos ^
sin fi — (l/L) sin a
cos fi = Vl  (WL 2 ) sin2 a
iin a 1 Z sin a
Vl  sin2« 2 VL2  J2 sin 2 a
Dividing by sin a and squaring both sides, we find
and from (11)
as required.
sin/? =
V4L2
Z2
iVs
V4L2
/2
Ly/Z
(10)
(«)
(12)
(IS)
(U)
(IS)
7.31. A uniform solid consists of a cylinder of radius a
and height H on a hemisphere of radius a, as indi
cated in Fig. 717. Prove that the solid is in stable
equilibrium on a horizontal plane if and only if
a/H > \/2.
By Problem 7.28 the centroid C is at a distance CB
from the center B of the hemisphere given by
„ _ 3q2 + SaH + 6ff 2 _ 6ff2  3a 2
8a + 12H ~ 8a + 12H
Then the distance of the centroid C above the plane is
CP = CD + DP = CB cos e + BQ
6W  3a2
8a + 12H
cos + a
186
SYSTEMS OF PARTICLES
[CHAP. 7
so that the potential energy (or potential) is
V
n* i 6ff 2  3a 2
tp ,.v • dV /3a 2 6H2\ .
Equilibrium occurs where — = or M# ( = — „ ) sin 9 — 0, i.e. = 0.
Then the equilibrium will be stable if
'3a 2  6# 2
aW
do 2
= M^
,=o v 8a + 12H
i.e. 3a 2  6H 2 > or a/H > V2.
cos 9
0=0
M#
3a 2  6ff 2 \
8a + 12H
>
7.32. A uniform chain has its ends suspended
from two fixed points at the same hori
zontal level. Find an equation for the
curve in which it hangs.
Let A and B [Fig. 718] be the fixed points.
An element of the chain of length As is in equilib
rium under the tensions of magnitude T and
T + AT due to the rest of the chain and also
the weight ag As of the element of chain. Now
from Fig. 718 if the directions of the vectors
corresponding to T and T + AT make angles
of 9 and 9 + A9 with the x axis respectively,
we have as the condition for equilibrium [neg
lecting terms of order (A9) 2 and higher], Fig. 718
(T + AT) cos (9 + A9)i + (T + AT) sin (9 + A9)j  (T cos 9 i + T sin 9 j)  ag] As
or (T + AT) cos {9 + A9) = T cos 9
(T + AT) sin (9 + A9)  T sin 9 = ag As
A
.B
//T +
A A e + Ae
A*l
f —
J^*r
agAs
*T ,,
/,
J
i
=
(1)
(2)
Equation (1) shows that the horizontal component T cos 9 must be a constant, which we shall take
as T which corresponds to the tension at the lowest point of the chain, where 9 = 0. Thus
Tcos9 = T n {3)
From (2) we find on dividing by A9,
(T + AT) sin(fl + Afl)
T sin9
A9
Taking the limit of both sides of U) as A9 * 0, we find
d
As
° 9 A~9
d9
(T sin 9) = ag
ds
d9
Using («?) to eliminate T, (5) becomes
d
j e (T tan 9)
T n
ds
° 9 T9
or
where b = Tjag. Now
Thus from (7) and (*),
dx
d9
dy
de
do
dx
de
dx ds
ds d9
dy ds
ds d9
ag
= COS 9,
sec 2 9
b sec 2 9
dy ■ n
f = sm9
ds
= (cos 9)(b sec 2 9)
= (sin 9){b sec 2 9)
= b sec 9
= b sec 9 tan 9
tt)
(5)
(6)
(7)
(8)
(9)
(10)
CHAP. 7] SYSTEMS OF PARTICLES 187
Integrating (9) and (10) with respect to $, we find
05 = 6 In (sec 6 + tan e) + Cj (11)
y = b sec $ + c 2 (12)
Let us assume that at the lowest point of the chain, i.e. at e = 0, x = and y = b. Then
from (11) and (12) we find Cj = 0, c 2 = 0. Thus
x = b In (sec + tan e) (IS)
y = b sec ff (14)
From (15) we have sec 6 + tan = e* /b (15)
But sec 2 e — tan 2 e = (sec + tan tf)(sec — tan e) = 1 (10)
Then dividing (1£) by (15), we find
sec e — tan 6 = e~*/ b (17)
Adding (15) to (17") and using (14), we find
This curve is called a catenary [from the Latin, meaning chain]
y = (ex/b + e */b) = 6 cosh  (15)
Supplementary Problems
DEGREES OF FREEDOM
7.33. Determine the number of degrees of freedom in each of the following cases: (a) a particle moving
on a plane curve; (6) two particles moving on a space curve and having constant distance between
them; (c) three particles moving in space so that the distance between any two of them is
always constant. Ans. (a) 1, (b) 1, (c) 6
7.34. Find the number of degrees of freedom for a rigid body which (a) moves parallel to a fixed
plane, (6) has two points fixed but can otherwise move freely. Ans. (a) 3, (6) 1
7.35. Find the number of degrees of freedom for a system consisting of a thin rigid rod which can
move freely in space and a particle which is constrained to move on the rod. Ans. 4
CENTER OF MASS AND MOMENTUM OF A SYSTEM OF PARTICLES
7.36. A quadrilateral ABCD has masses 1, 2, 3 and 4 units located at its vertices A(— 1,2,2),
B(S, 2, 1), C(l, —2, 4) and D(3, 1, 2). Find the coordinates of the center of mass. Ans. (2, 0, 2)
7.37. A system consists of two particles of masses m x and m 2 . Prove that the center of mass of the
system divides the line joining Wj to m 2 into two segments whose lengths are in the ratio m? to Wj.
7.38. A bomb dropped from an airplane explodes in midair. Prove that if air resistance is neglected,
then the center of mass describes a parabola.
7.39. Three particles of masses 2, 1, 3 respectively have position vectors r t = 5ti — 2t 2 j + (St — 2)k,
r 2 = (2t  3)i + (12  5t 2 )j + (4 + 6*  3*3)k, r 3 = (2*  1)1 + (« 2 + 2)j  tf» k where t is the time.
Find (a) the velocity of the center of mass at time t = 1 and (6) the total linear momentum
of the system at t = 1. Ans. (a) 3i — 2j — k, (6) 181 — 12j — 6k
188
SYSTEMS OF PARTICLES
[CHAP. 7
7.40. Three equal masses are located at the vertices of a triangle. Prove that the center of mass is
located at the intersection of the medians of the triangle.
7.41. A uniform plate has the shape of the region bounded by the parabola
y = x 2 and the line y — H in the xy plane. Find the center of mass.
Ans. x = 0, y = ^H
7.42. Find the center of mass of a uniform right circular cone of radius a
and height H.
Ans. That point on the axis at distance £i? from the vertex.
7.43. The shaded region of Fig. 719 is a solid spherical cap of height H
cut off from a uniform solid sphere of radius a. (a) Prove that the
centroid of the cap is located at a distance f (2a — H) 2 /(Sa — H) from
the base AB. (b) Discuss the cases H — 0, H — a and H = 2a.
7.44. Find the center of mass of a uniform plate bounded by
y = sin x and the x axis. Ans. x — tt/2, y — ir/&
7.45. Find the center of mass of a rod of length I whose den
sity is proportional to the distance from one end O.
Ans. Z from end O
7.46. Find the centroid of a uniform solid bounded by the
planes Ax + 2y + z = 8, x — 0, y = 0, 2 = 0.
Ans. f = j^«(i + 2j + 4k)
7.47. A uniform solid is bounded by the paraboloid of revolu
tion x 2 + y 2 = cz and the plane z = H [see Fig. 720].
Find the centroid. Ans. x = 0, y = 0, z = %H
Fig. 719
Fig. 720
ANGULAR MOMENTUM AND TORQUE
7.48. Three particles of masses 2, 3 and 5 move under the influence of a force field so that their
position vectors relative to a fixed coordinate system are given respectively by r x = 2ti — 3j + t 2 k,
r 3 = (t + l)i + 3tj  4k and r 3 = < 2 i — *j + (2t  l)k where t is the time. Find (a) the total
angular momentum of the system and (b) the total external torque applied to the system, taken
with respect to the origin.
Ans. (a) (31  12t)i + (6t 2  lOt  12)j + (21 + 5* 2 )k
(b) 12i + (12*  10)j + lOtk
7.49. Work Problem 7.48 if the total angular momentum and torque are taken with respect to the
center of mass.
7.50. Verify that in (a) Problem 7.48 and (6) Problem 7.49 the total external torque is equal to the
time rate of change in angular momentum.
7.51. In Problem 7.48 find (a) the total angular momentum and (b) the total external torque taken
about a point whose position vector is given by r = ti  2tj + 3k. Does the total external torque
equal the time rate of change in angular momentum in this case? Explain.
7.52. Verify Theorem 7.9, page 169, for the system of particles of Problem 7.48.
7.53. State and prove a theorem analogous to that of Theorem 7.9, page 169, for the total external
torque applied to a system.
7.54. Is the angular momentum conserved in Problem 7.38? Explain.
CHAP. 7]
SYSTEMS OF PARTICLES
189
WORK, ENERGY AND IMPULSE
7.55. Find the total work done by the force field of Problem 7.48 in moving the particles from their
positions at time t — 1 to their positions at time t = 2. Ans. 42
7.56. Is the work of Problem 7.55 the same as that done on the center of mass assuming all mass
to be concentrated there? Explain.
7.57. Find the total kinetic energy of the particles in Problem 7.48 at times (a) t = 1 and (b) t — 2.
Discuss the connection between your results and the result of Problem 7.55.
Ans. (a) 72.5, (6) 30.5
7.58. Find the total linear momentum of the system of particles in Problem 7.48 at times (a) t = 1 and
(6) t = 2. Ans. (a) 17i + 4j + 14k, (6) 27i + 4j + 18k
7.59. Find the total impulse applied to the system of Problem 7.48 from t = 1 to t  2 and discuss
the connection of your result with Problem 7.58. Ans. lOi + 4k
7.60. Prove Theorem 7.13, page 170.
7.61. Verify Theorem 7.13, page 170, for the system of particles in Problem 7.48.
CONSTRAINTS, STATICS, VIRTUAL WORK, STABILITY AND D'ALEMBERT'S PRINCIPLE
7.62. In each case state whether the constraint is holonomic or nonholonomic and give a reason for
your answer: (a) a particle constrained to move under gravity on the inside of a vertical paraboloid
of revolution whose vertex is downward; (6) a particle sliding on an ellipsoid under the influence
of gravity; (c) a sphere rolling and possibly sliding down an inclined plane; (d) a sphere rolling
down an inclined plane parallel to a fixed vertical plane; (e) a particle sliding under gravity on
the outside of an inverted vertical cone.
Ans. (a) holonomic, (6) nonholonomic, (c) nonholonomic, (d) holonomic, (e) holonomic
7.63. A lever ABC [Fig. 721] has weights W t and W 2
at distances a t and a 2 from the fixed support B.
Using the principle of virtual work, prove that a
necessary and sufficient condition for equilibrium is
W t a x  W 2 a 2 .
7.64. Work Problem 7.63 if one or more additional weights
are placed on the lever.
7.65. An inextensible string of negligible mass hanging
over a smooth peg at B [see Fig. 722] connects one mass
m x on a frictionless inclined plane of angle a to another
mass m 2 . Using D'Alembert's principle, prove that
the masses will be in equilibrium if m 2 = m 1 sin a.
7.66. Work Problem 7.65 if the incline has coefficient of fric
tion ft. Ans. m 2 = m^sin a — fi cos a)
W*
B
IE
Og
Fig. 721
W*
Fig. 722
7.67. A ladder AB of mass m has its ends on a smooth wall and floor
[see Fig. 723]. The foot of the ladder is tied by an inextensible
rope of negligible mass to the base C of the wall so that the
ladder makes an angle a with the floor. Using the principle
of virtual work, find the magnitude of the tension in the rope.
Ans. \mg cot a
7.68. Work (a) Problem 7.63 and (6) Problem 7.65 by using the po
tential energy method. Prove that the equilibrium in each case
is unstable.
Fig. 7
190
SYSTEMS OF PARTICLES
[CHAP. 7
7.69. A thin uniform rod of length I has its two ends constrained to move on
the circumference of a smooth vertical circle of radius a [see Fig. 724].
Determine conditions for equilibrium.
7.70. Is the equilibrium of the rod of Problem 7.69 stable or not? Explain.
7.71. A solid hemisphere of radius a is located on a perfectly rough inclined
plane of angle a.
(a) Prove that it is in stable equilibrium if a < sin 1 (3/8).
(b) Are there any other values of a for which equilibrium can occur?
Which of these, if any, yield stable equilibrium? Fig. 724
7.72. Use D'Alembert's principle to obtain the equations of motion of masses m 1 and w 2 of Problem 7.65.
7.73. Work Atwood's machine problem [see Problem 7.22, page 180] by using D'Alembert's principle.
7.74. Use D'Alembert's principle to determine the equations of motion of a simple pendulum.
MISCELLANEOUS PROBLEMS
7.75. Prove that the center of mass of a uniform circular arc of radius a and central angle a is
located on the axis of symmetry at a distance from the center equal to (a sin a)/a.
7.76. Discuss the cases (a) a = tt/2 and (6) a = w in Problem 7.75.
7.77.
7.78.
7.79.
7.80.
7.81.
7.82.
7.83.
7.84.
7.85.
A circle of radius a is removed from a uniform circular plate
of radius b > a, as indicated in Fig. 725. If the distance be
tween their centers A and B is D, find the center of mass.
Ans. The point at distance a 2 D/(b 2  a 2 ) below B.
Work Problem 7.77 if the circles are replaced by spheres.
Ans. The point at distance a 3 D/(b 3  a 3 ) below B.
Prove that the center of mass does not depend on the origin
of the coordinate system used.
Prove that the center of mass of a uniform thin hemispherical
shell of radius a is located at a distance £a from the center.
Fig. 725
Let the angular momentum of the moon about the earth be A. Find the angular momentum of a
system consisting of only the earth and the moon about the center of mass. Assume the masses
of the earth and moon to be given by M e and M m respectively. Ans. M e A/(M e + M m )
Does Theorem 7.13, page 170, apply in case the angular momentum is taken about an arbitrary
point? Explain.
In Fig. 726, AD, BD and CD are uniform thin rods
of equal length a and equal weight w. They are
smoothly hinged at D and have ends A, B and C on a
smooth horizontal plane. To prevent the motion
of ends A, B and C, we use an inextensible string
ABC of negligible mass which is in the form of an
equilateral triangle. If a weight W is suspended
from D so that the rods make equal angles a with
the horizontal plane, prove that the magnitude of
the tension in the string is %V% (W + 3w) cot a.
Work Problem 7.83 if the weight W is removed
from D and suspended from the center of one of
the rods.
Fig. 726
Derive an expression for (a) the total angular momentum and (b) the total torque of a system about
an arbitrary point.
CHAP. 7]
SYSTEMS OF PARTICLES
191
7.86. Prove that the torque about any point P is equal to the time rate of
change in angular momentum about P if and only if (a) P is fixed in
space, (b) P coincides with the center of mass or (c) P is moving with
a velocity which is in the same direction as the center of mass.
7.87. Find the centroid of a solid of constant density consisting of a right
circular cone of radius a and height H surmounted by a hemisphere of
radius a [see Fig. 727].
Ans. At height £(a 2 + H 2 )/(2a + H) above O.
7.88. Work Problem 7.87 if the density of the cone is twice the density of
the hemisphere. Ans. At height § (a 2 + 2H 2 )/(a + H) above O.
7.89. A hemisphere of radius a is cut out of a uniform solid cube of side b > 2a [see Fig. 728]. Find
the center of mass of the remaining solid.
Fig. 728
Fig. 729
Fig. 730
7.90. A uniform chain of 45kgwt is suspended from two fixed supports 15 meters apart. If the sag
in the middle is 20 cm, find the tension at the supports. Ans. 450 kg wt
7.91. A chain of length L and constant density a is suspended from two fixed points at the same
horizontal level. If the sag of the chain at the middle is at a distance D below the horizontal
line through the fixed points, prove that the tension at the lowest point of the chain is
a(L2  4D 2 )/8D.
7.92. Three particles of masses m l ,m 2 ,m 3 are located at the vertices of a triangle opposite sides having
lengths a lt a 2 ,a 3 respectively. Prove that the center of mass lies at the intersection of the angle
bisectors of the triangle if and only if mj/aj = m 2 /a 2 = m 3 /a 3 .
7.93. Masses m x and m 2 are on a frictionless circular cylinder connected by an inextensible string of
negligible mass [see Fig. 729]. (a) Using the principle of virtual work, prove that the system
is in equilibrium if m 1 sin ai = m 2 sin a 2 . (b) Is the equilibrium stable? Explain.
7.94. Work Problem 7.93 if friction is taken into account.
7.95. Derive an expression for the total kinetic energy of a system of particles relative to a point
which may be moving in space. Under what conditions is the expression mathematically
simplified? Discuss the physical significance of the simplification.
7.96. Find the center of mass of a uniform plate shown shaded in Fig. 730 which is bounded by the
hypocycloid x 2 ' z + y 2 ' s = a 2 ' 3 and the lines x = 0, y = 0. [Hint. Parametric equations for the
hypocycloid are x — a cos 3 e, y = a sin 3 e.] Ans. x — y = 256a/315jr
192
SYSTEMS OF PARTICLES
[CHAP. 7
7.97. Let m 1( m 2 , m 3 be the masses of three particles and v 12 , v 23 , v 13 be their relative velocities.
(a) Prove that the total kinetic energy of the system about the center of mass is
,2
m x m 2 v\ 2 + m 2 m 3 V23 + m 1 m i v'x Z
m x + m 2 + m 3
(6) Generalize the result in (a).
7.98. A chain of variable density is suspended from two fixed points on the same horizontal level.
Prove that if the density of the chain varies as the horizontal distance from a vertical line
through its center, then the shape of the chain will be a parabola.
7.99. Discuss the relationship of Problem 7.98 with the shape of a suspension bridge.
7.100. A solid consists of a uniform right circular cone of vertex angle a on a uniform hemisphere of
the same density, as indicated in Fig. 731. Prove that the solid can be in stable equilibrium on
a horizontal plane if and only if a > 60°.
Fig. 731
Fig. 732
7.101. A uniform solid [see Fig. 732] consists of a hemisphere of radius a surmounted by a cube of
side b symmetrically placed about the center of the hemisphere. Find the condition on a and b
for stable equilibrium.
Ans. alb > yWv
7.102. Find the centroid of the area bounded by the cycloid
x = a(e — sin e), y = a(l — cos e)
and the x axis. Ans. (ira, 5a/6)
7.103. Prove that if the component of the torque about point P in any direction is zero, then the
component of angular momentum about P in that direction is conserved if (a) P is a fixed point,
(6) P coincides with the center of mass or (c) P is a point moving in the same direction as
the center of mass.
7.104. In Problem 7.103, is the angular momentum conserved only if (a), (6) or (c) occurs? Explain.
7.105. Prove that the virtual work due to a force is equal to the sum of the virtual works which cor
respond to all components of the force.
7.106. Prove that it is impossible for one sphere to be in stable equilibrium on a fixed sphere which is
perfectly rough [i.e. with coefficient of friction p = 1]. Is it possible for equilibrium to occur
at all? Explain.
7.107. A uniform solid having the shape of the paraboloid of revolution cz = x 2 + y 2 , c > rests on
the xy plane, assumed horizontal. Prove that if the height of the paraboloid is H, then the
equilibrium is stable if and only if H < f c.
7.108. Work Problem 7.107 if the xy plane is inclined at an angle a with the horizontal.
CHAP. 7] SYSTEMS OF PARTICLES 193
7.109. In Fig. 733, AC and BC are frictionless wires in a vertical plane making angles of 60° and 30°
respectively with the horizontal. Two beads of masses 3 gm and 6 gm are located on the wires,
connected by a thin rod of negligible mass. Prove that the system will be in equilibrium
when the rod makes an angle with the horizontal given by tan 1 (#y3 )•
Fig. 733
7.110. Prove each of the following theorems due to Pappus.
(a) If a closed curve C in a plane is revolved about an axis in the plane which does not intersect
it, then the volume generated is equal to the area bounded by C multiplied by the distance
traveled by the centroid of the area.
(b) If an arc of a plane curve (closed or not) is revolved about an axis in the plane which does
not intersect it, then the area of the surface generated is equal to the length of the arc
multiplied by the distance traveled by the centroid of the arc.
7.111. Use Pappus' theorems to find (a) the centroid of a semicircular plate, (b) the centroid of a semi
circular wire, (c) the centroid of a plate in the form of a right triangle, (d) the volume of a cylinder.
7.112. Find the (a) surface area and (6) volume of the doughnut shaped region obtained by revolving
a circle of radius a about a line in its plane at a distance b > a from its center.
Ans. (a) A^ab, (b) l^a^b
Chapter 8 APPLICATIONS to
VIBRATING SYSTEMS,
ROCKETS and COLLISIONS
VIBRATING SYSTEMS OF PARTICLES
If two or more particles are connected by springs [or interact with each other in some
equivalent manner], then the particles will vibrate or oscillate with respect to each other.
As seen in Chapter 4, a vibrating or oscillating particle such as the simple harmonic
oscillator or bob of a simple pendulum, has a single frequency of vibration. In the case
of systems of particles, there is generally more than one frequency of vibration. Such
frequencies are called normal frequencies. The motions of the particles in these cases are
often called multiply periodic vibrations.
A mode of vibration [i.e. a particular way in which vibration occurs, due to particular
initial conditions for example] in which only one of the normal frequencies is present is called
a normal mode of vibration or simply a normal mode. See Problems 8.18.3.
PROBLEMS INVOLVING CHANGING MASS. ROCKETS
Thus far we have restricted ourselves to motions of particles having constant mass.
An important class of problems involves changing mass. An example is that of a rocket
which moves forward by expelling particles of a fuel mixture backward. See Problems
8.4 and 8.5.
COLLISIONS OF PARTICLES
During the course of their motions two or more particles may collide with each other.
Problems which consider the motions of such particles are called collision or impact
problems.
In practice we think of colliding objects, such as spheres, as having elasticity. The
time during which such objects are in contact is composed of a compression time during
which slight deformation may take place, and restitution time during which the shape is
restored. We assume that the spheres are smooth so that forces exerted are along the
common normal to the spheres through the point of contact [and passing through their
centers] .
A collision can be direct or oblique. In a direct collision the direction of motion of both
spheres is along the common normal at the point of contact both before and after collision.
A collision which is not direct is called oblique.
Fundamental in collision problems is the following principle called Newton's collision
rule which is based on experimental evidence. We shall take it as a postulate.
Newton's collision rule. Let v 12 and v^ be the relative velocities of the spheres
along the common normal before and after impact. Then
V 12 ~ ~~ eV 12
194
CHAP. 8] APPLICATIONS TO VIBRATING SYSTEMS, ROCKETS AND COLLISIONS
195
The quantity c, called the coefficient of restitution, depends on the materials of which the
objects are made and is generally taken as a constant between and 1. If e = the
collision is called perfectly inelastic or briefly inelastic. If € = 1 the collision is called
perfectly elastic or briefly elastic.
In the case of perfectly elastic collisions the total kinetic energy before and after impact
is the same.
CONTINUOUS SYSTEMS OF PARTICLES
For some problems the number of particles per unit length, area or volume is so large
that for all practical purposes the system can be considered as continuous. Examples are a
vibrating violin string, a vibrating drumhead or membrane, or a sphere rolling down an
inclined plane.
The basic laws of Chapter 7 hold for such continuous systems of particles. In applying
them, however, it is necessary to use integration in place of summation over the number
of particles and the concept of density.
THE VIBRATING STRING
Let us consider an elastic string such as a violin or piano string which is tightly
stretched between the fixed points x = and x = I of the x axis [see Fig. 81]. If the
string is given some initial displacement [such as, for example, by plucking it] and is then
released, it will vibrate or oscillate about the equilibrium position.
x =
x = I
x = l
Fig. 81
Pig. 82
If we let Y(x, t) denote the displacement of any point x of the string from the equilibrium
position at time t [see Fig. 82], then the equation governing the vibrations is given by the
partial differential equation
d 2 Y n d 2 Y
dt 2
= c<
dx 2
where if T is the (constant) tension throughout the string and a is the (constant) density
[mass per unit length of string],
= Tfa
(2)
The equation (1) holds in case the vibrations are assumed so small that the slope dY/dx
at any point of the string is much less than one.
BOUNDARYVALUE PROBLEMS
The problem of solving an equation such as (1) subject to various conditions, called
boundary conditions, is often called a boundaryvalue problem. An important method
for solving such problems makes use of Fourier series.
196
APPLICATIONS TO VIBRATING SYSTEMS, ROCKETS AND COLLISIONS [CHAP. 8
FOURIER SERIES
Under certain conditions [usually satisfied in practice and outlined below] a function
f(x), defined in the interval y < x < y + 21 and having period 21 outside of this interval,
has the series expansion
f\ x ) = o" ^ ( ttn cos ~~ l — sm ~T~ )
where the coefficients in the series, called Fourier coefficients, are given by
,y + 2l
j s*y + 2l
nirX .
cos — r  aa?
# sm
7
dx
(3)
(5)
Such a series is called the Fourier series of /(a;). For many problems y = or —I.
ODD AND EVEN FUNCTIONS
If y = — I, certain simplifications can occur in the coefficients (4) and (5) as indicated
below:
1. If f(x) = f(x),
a n = j I /(a;) cos — j— aaj,
Z
bn =
(6)
In such case f(x) is called an even function and the Fourier series corresponding to
f(x) has only cosine terms.
2. If f(x) = f(x),
a n = 0,
6 n
fX'^>
sin — =— ax
{7)
In such case f(x) is called an odd function and the Fourier series corresponding to f(x)
has only sine terms.
If f(x) is neither even nor odd its Fourier series will contain both sine and cosine terms.
Examples of even functions are x 4 , Sx 6 + 4x 2 — 5, cos a;, e x + e~ x and the function shown
graphically in Fig. 83. Examples of odd functions are x 3 ,2x 5 — 5x 3 + 4, sinx, e x — e~ x and
the function shown graphically in Fig. 84.
Examples of functions which are neither even nor odd are x 4 + x 3 , x + cos x and the
function shown graphically in Fig. 85.
fix)
Fig. 83
Fig. 84
Fig. 85
If a function is defined in the "half period" x = to x = I and is specified as odd,
then the function is known throughout the interval — I < x < I and so the series which
CHAP. 8] APPLICATIONS TO VIBRATING SYSTEMS, ROCKETS AND COLLISIONS
197
contains only sine terms can be found. This series is often called the half range Fourier
sine series. Similarly, a function denned from x = to x = I which is specified as even
has a series expansion called the half range Fourier cosine series.
CONVERGENCE OF FOURIER SERIES
Let us assume the following conditions on f(x):
1. f{x) is defined in y < x < y + 21. H x)
/(*i + 0)
f(x) and its derivative f'(x) are piece
wise continuous in y < x < y + 21. [A
function is said to be piecewise contin
uous in an interval if the interval can
be divided into a finite number of sub
intervals in each of which the function
is continuous and bounded, i.e. there
is a constant B > such that
—B < f(x) < B. An example of such a
function is indicated in Fig. 86.]
/(* 2 +0)
Fig. 86
3. At each point of discontinuity, for example, Xi [or x 2 ] in Fig. 86, f(x) has finite limits
from the right and left denoted respectively by f(xi + 0) and f{xi — 0) [or f(x 2 + 0),
f(X2~0)}.
4. f(x) has period 21, i.e. f(x + 21) = f(x).
These conditions if satisfied are sufficient to guarantee the validity of equation (3)
[i.e. the series on the right side of (3) actually converges to f(x)) at each point where f(x)
is continuous. At each point where f(x) is discontinuous, (3) is still valid if f(x) is replaced
by £[/(# + 0) 4 f(x — 0)], i.e. the mean value of the right and left hand limits.
The conditions described above are known as Dirichlet conditions.
Solved Problems
VIBRATING SYSTEMS OF PARTICLES
8.1. Two equal masses m are connected by
springs having equal spring constant k,
as shown in Fig. 87, so that the masses
are free to slide on a frictionless table
AB. The walls at A and B to which the
ends of the springs are attached are fixed.
Set up the differential equations of
motion of the masses.
Let x x i and x 2 i [Fig. 88] denote the dis
placements of the masses from their equilibrium
positions C and D at any time t.
••" nit
T5WKT ( J— ^RRKP C ,#"
OH5ffir>
Fig. 87
C ni
^1 *ii u \ x 2 i
nmu^
"ChJ ^^m^ — £Lj r ww^
P
Fig. 88
Q
198 APPLICATIONS TO VIBRATING SYSTEMS, ROCKETS AND COLLISIONS [CHAP. 8
Consider the forces acting on the first mass at P. There will be a force due to the spring on
the right given by K (x 2 i — a^i) = K (x 2  a^i, and a force due to the spring on the left given by
— KXyi. Thus the net force acting on the first mass at P is
k(x 2 — x t )i — KX t i
In the same way the net force acting on the second mass at Q is
k(x x — x 2 )i — kx 2 i
Then by Newton's second law we have
d 2
mr^(x x \) — k(x 2 — x 1 )i — KX t i
d 2
or m 'x\ = k(x 2 ~ 2Xi) (l)
m x 2 = k(x 1 — 2x 2 ) (2)
8.2. Find (a) the normal frequencies and (b) the normal modes of vibration for the
system in Problem 8.1.
(a) Let x 1 = A t cos at, x 2 = A 2 cosut in equations (1) and (2) of Problem 8.1. Then we find,
after simplifying,
(2Kmu 2 )A 1  K A 2 = (1)
icA 1 + (2/cm» 2 )A 2 = (2)
Now if Ay and A 2 are not both zero, we must have
2k — ffllO 2 — K
2 = ° <*>
— jc 2k — Ww 2
(2/c — ?ww 2 )(2»c — mw 2 ) — k 2 = or m 2 w 4 — 4/cra<o 2 + 3/c 2 =
Aktyi ± \/16k 2 w 2 — 12/c 2 m 2
Solving for w 2 , we find <o 2 = — = giving
(o 2 = /c/m and w 2 = 3/c/m (.4)
Then the normal (or natural) frequencies of the system are given by
» 1 / k . 1 / 3/c
/ = 7T \\~ and / = — \\ —
2v \ m 2tt \ m
(5)
The normal frequencies are also called characteristic frequencies and the determinant (5)
is called the characteristic determinant or secular determinant.
(b) To find the normal mode corresponding to u = yficfm, let w 2 = ic/m in equations (i) and (2).
Then we find
A t = A 2
In this case the normal mode of vibration corresponds to the motion of the masses in the same
direction [i.e. both to the right and both to the left] as indicated in Fig. 89.
Normal mode corresponding to w = yj nlm Normal mode corresponding to w = yZic/m
Fig. 89 Fig. 810
Similarly to find the normal mode corresponding to w = ySK/m, let « 2 = 3ic/m in equations
(1) and («). Then we find
A x — —A 2
CHAP. 8] APPLICATIONS TO VIBRATING SYSTEMS, ROCKETS AND COLLISIONS 199
In this case the normal mode of vibration corresponds to the motion of the masses in opposite
directions [i.e. when one moves to the right the other moves to the left, and vice versa] as
indicated in Fig. 810 above.
In working this problem we could also just as well have assumed, x t = B x sin at,
x 2 = B 2 sin a t or x x = A x cos at + B x sin at, x 2 = A 2 cos at + B 2 sin at or x x = C x e™*,
x 2  C 2 e ibit .
8.3. Suppose that in Problem 8.1 the first mass is held at its equilibrium position while
the second mass is given a displacement of magnitude a > to the right of its
equilibrium position. The masses are then released. Find the position of each mass
at any later time.
Writing Wl = ^f^Jm and « 2 = V^c/w, the general motion of both masses is described by
x x  C 1 cosu 1 t + C 2 sin ujt + C 3 cos<o 2 £ + C 4 sin w 2 i (0
x 2 = D x coswx* + D 2 sinwjt + D 3 cos u 2 * + D± sin a 2 t (2)
where the coefficients are all constants. Substituting these into equation (1) or equation (2) of
Problem 8.1 [both give the same results], we find on equating corresponding coefficients of cos Wl t,
sin a^t, cos u 2 t, sin a 2 t respectively,
Dj = C lf D 2 = C 2 , D 3 = C 3 , D 4 = C 4
Thus equations (1) and (2) can be written
x x — C x cos o^t + C 2 sin u x t + C 3 cos <o 2 < + C 4 sin a 2 t (3)
x 2 = Cxcosuit + C 2 sin u x t — C 3 cosu 2 t ~ C 4 sina 2 t (4)
We now determine Cj,C 2 , C 3 , C 4 subject to the initial conditions
Xl = 0, x 2 = a, x x  0, x 2 — at t = (5)
From these conditions we find respectively
C 1 + C 3 = 0, C x  C 3 = a, C 2 W! + C 4 <o 2 = 0, C 2 o^  C 4 w 2 =
From these we find C x = \a, C 2 = 0, C 3 = ia, C 4 = («)
Thus equations (5) and (4) give the required equations
x i — ^«(cos w x t — cos o> 2 t) (7)
x 2 — ^a(cos u^t + cos o) 2 t) {8)
where a x = V '/c/w, w 2 = v 3*/m.
Note that in the motion described by (7) and (8), both normal frequencies are present. These
equations show that the general motion is a superposition of the normal modes. This is sometimes
called the superposition principle.
CHANGING MASS. ROCKETS
8.4. Derive an equation for the motion of a v + v v + Av
rocket moving in a straight line.
o 1 , o —
—Aw m + Am
Let m be the total mass of the rocket at q
time t. At a later time t + At suppose that
the mass is m + Am due to expulsion of a mass
—Am of gas through the back of the rocket.
Note that —Am is actually a positive quantity Fig. 811
since Am is assumed negative.
Let v and v + Av be the velocities of the rocket at times t and t + At relative to an inertial
system with origin at O. The velocity of the mass of gas ejected from the back of the rocket relative
to O is v + v where — v is the velocity of the gas relative to the rocket.
200
APPLICATIONS TO VIBRATING SYSTEMS, ROCKETS AND COLLISIONS [CHAP. 8
Since the change in momentum of the system is equal to the impulse, we have
Total momentum at t + At — total momentum at t = impulse
{(ra + Am)(v + Av) + (Am)(v + v )}  mv = F At (l)
where F is the net external force acting on the rocket.
Equation (1) can be written as
Av Am Av
m M ~ V °^ + AT Am = F
Then taking the limit as At > 0, we find
dv dm
m di ~ v °dT = F W
Writing v = vi, v = v i, F = Fi, this becomes
dv . dm _
m Tt + v ° It = F W
8.5. Find the velocity of the rocket of Problem 8.4 assuming that gas is ejected at a
constant rate and at constant velocity with respect to it and that it moves vertically
upward in a constant gravitational field.
If the gas is ejected at constant rate a > 0, then m = m at where m is the mass of the
rocket at t  0. Since F = mg\ (or F  mg) and dm/dt = a, equation (S) of Problem 8.4
can be written
/ s\ dv . dv aV o
(m at) —  a v = (m  a t)g or ^ =  9 + ——^ {1)
Integrating, we find v = gt  v In (ra  at) + c x (2)
If v = at t = 0, i.e. if the rocket starts from rest, then
=  v In m + c x or c x = v In m
Thus (2) becomes v  —gt + v n \n ( — — — J
\m  at J
which is the speed at any time. The velocity is v = vi.
Note that we must have m  at > 0, otherwise there will be no gas expelled from the rocket,
in which case the rocket will be out of fuel.
COLLISIONS OF PARTICLES
8.6. Two masses mi and m 2 traveling in the same straight line collide. Find the velocities
of the particles after collision in terms of the velocities before collision.
Assume that the straight line is taken to be
the x axis and that the velocities of the particles
before and after collisions are v,,v 9 and vi, v,
respectively. 1 /~ \ v * » v v v 2 ^
By Newton's collision rule, page 194,
▼iv£ = e(v 2  Vl ) (1) Fig. 812
By the principle of conservation of momentum,
Total momentum after collision = total momentum before collision
m x \[ + m 2 V2 = m 1 \ 1 + m 2 v 2 (2)
CHAP. 8] APPLICATIONS TO VIBRATING SYSTEMS, ROCKETS AND COLLISIONS
201
Solving (1) and (2) simultaneously,
V2 =
(m x  em 2 )v 1 + m 2 (l + e)v 2
fltj + W&2
m 1 (l + e)vj + (m 2 — «w 1 )v 2
m t + m 2
(S)
8.7. Discuss Problem 8.6 for the case of (a) a perfectly inelastic collision, (b) a perfectly
elastic collision.
(a) Here we put e = in (3) and (4) of Problem 8.6 to obtain
vi
m^Vi + m 2 v 2
Wj + ra 2 '
v 2 =
m^ + m2V 2
mj + m 2
Thus after collision the two particles move with the same velocity, i.e. they move as if they
were stuck together as a single particle.
(b) Here we put e — 1 in (3) and (4) of Problem 8.6 to obtain
(ra x — m 2 )v 1 + 2m 2 v 2 , 2m l \ l + (mj — m^v 2
v i = ^ j,^ » v 2 = 
wii + m 2
These velocities are not the same.
m, + w 2
8.8. Show that for a perfectly elastic collision of the particles of Problem 8.6 the total
kinetic energy before collision equals the total kinetic energy after collision.
Using the result of Problem 8.7(&), we have
Total kinetic energy after collision = n w i v i 2 + ~m 2 v 2 2
= 2 mi
(m x
m 2 )v! + 2w 2 v 2 l 2
1 2 , 1 z
m x + m 2
2
+ 2%
C2m l v 1 + (w 2 — m!)v 2 ] 2
= total kinetic energy before collision
8.9.
Two spheres of masses mi and ra 2 respectively, collide obliquely. Find their velocities
after impact in terms of their velocities before impact.
Let v 1 ,v 2 and vi,v 2 be the velocities of the
spheres before and after impact respectively, as
indicated in Fig. 813. Choose a coordinate system
so that the xy plane is the plane of v t and v 2 , and
so that at the instant of impact the x axis passes
through the centers C t and C 2 of the spheres.
By the conservation of momentum, we have
m 1 v 1 + m 2 v 2 = w^vi + wi 2 v 2
From Fig. 813 we see that
(1)
Fig. 813
v i = ^iCcos *i i — sin e x j)
v 2 = v 2 (cos *2 * — sin e % j)
v i — ^i(cos <Pi i — sin <f> t j)
v 2 = v 2 (cos 2 i — sin <f> 2 j)
Substituting equations (2)(5) in (1) and equating coefficients of i and j, we have
(3)
U)
(5)
202
APPLICATIONS TO VIBRATING SYSTEMS, ROCKETS AND COLLISIONS [CHAP. 8
ra^ cos e i + m 2 v 2 cos B 2 = m^ cos <p x + m 2 v' 2 cos 2
m 1 v 1 sin e l + m 2 v 2 sin e 2 = m^v'^ sin </>! + m 2 i> 2 sin <p 2
By Newton's collision rule, we have
Relative velocity after impact along x axis
= — e{relative velocity before impact along x axis}
or vi • i  v 2 • i = e(v! • i  v 2 • i)
which on using equations {2)(5) becomes
v[ cos 0! — v 2 cos <f> 2 = —e(v l cos t — v 2 cos e 2 )
Furthermore, since the tangential velocities before and after impact are equal,
Vi'j = vi'j
v 2 *3 = v 2 «j
or v x sin e l = v[ sin <f> t
v 2 sin e 2 — v'2 sm 02
Equation (7) is automatically satisfied by using equations (12) and (13).
From equations (6) and (9) we find
(wj — m 2 e)v 1 cos ! + m 2 (l + e)t> 2 cos e 2
V x COS 0! =
V 2 COS 2 =
Then using (12) and (15) we find
m 1 + m 2
m 1 (l + e)Vi cos J + (m 2 — wiie)^ cos #2
m l + m 2
v i = ^i(cos 0i i — sin X j)
(m l — m 2 e)v 1 cos X i + ra 2 (l + e)v 2 cos 2 i
~~ m t + m 2
v 2 = v 2 (cos 02 i _ sin $2 J)
m^l + e)i>! cos *! i + (m 2 — m 1 e)v 2 cos 2 i
— wij + m 2
t>i sin e x j
— t> 2 sin e 2 j
(7)
(*)
(»)
(10)
(J*)
(18)
CONTINUOUS SYSTEMS OF PARTICLES
8.10. Derive the partial differential equation (1), page 195, for the transverse vibrations
of a vibrating string.
x + Ax
Fig. 814
Let us consider the motion of an element of the string of length As, greatly magnified in
Fig. 814.
The forces acting on the element due to the remainder of the string are given by the tensions,
as shown in Fig. 814, of magnitude T(x) and T(x + A«) at the ends x and x + As of the element.
CHAP. 8] APPLICATIONS TO VIBRATING SYSTEMS, ROCKETS AND COLLISIONS 203
The net horizontal force in direction i acting on the element is
[T(x + Ax) cos e(x + Ax)  T(x) cos e(x)]i (1)
The net vertical force in direction j acting on the element is
[T(x + Ax) sin 6(x + Ax) — T(x) sin e(x)]j (2)
If we assume that the horizontal motion in direction i is negligible, the net force (1) is zero.
Using the fact that the acceleration of the element is d 2 Y/dt 2 approximately and that its mass is
a As where a is the mass per unit length, we have from (2) and Newton's second law,
d 2 Y
a As TT^ j = [T(x + Ax) sin 6 (x + Ax) — T(x) sin e(x)] j (3)
or, dividing by Ax j,
As d*Y _ T(x + Aas) sin 6{x + Ax)  T(x) sin 6(x)
G Ax dt 2 ~ Ax
or L  /AY\*d 2 Y _ T(x + Ax) sin e(x + As)  T(x) sin e(x)
a \ \Ax J dt 2 Ax
Taking the limit as Ax * 0, this becomes
' "d 2 Y d
U)
•>Rs)
Since sin 9 =
U 2  dx( Tsin ^ &
tan e dY/dx
Vl + tan2 e Vl + (dY/dx) 2
equation (5) can be written
I + (&Y\ 2 PY = _j_ f TdY/dx__
\ \8xJ dt 2 dx 1 Vl + (dY/dx)*
(6)
To simplify this equation we make the assumption that vibrations are small so that the slope dY/dx
is small in absolute value compared with 1. Then we can neglect (dY/dx) 2 compared with 1 and
W becomes
If we further assume that the tension T is constant throughout the string and that a is also
constant, (7) becomes
c2 idr (8)
d 2 Y 2 § 2 Y
dt 2 ~ C dx 2
where c 2 = T/a. Unless otherwise specified, when we deal with the vibrating string we shall
refer to equation (8).
8.11. Derive the equation of Problem 8.10 if the string is horizontal and gravity is taken
into account.
In this case we must add to the right hand side of equation (3) of Problem 8.10 the force
on the element due to gravity
— mg = —a As gj
The effect of this is to replace equation (8) of Problem 8.10 by
d 2 Y _ 2 &Y _
dt 2 C dx 2 9
FOURIER SERIES
8.12. Graph each of the following functions.
f 3 <x < 5
<«)/<*> = _3  5 <*<0 Peri °d = 10
204
APPLICATIONS TO VIBRATING SYSTEMS, ROCKETS AND COLLISIONS [CHAP. 8
/(*)
Per
3
iod
~*~
1
25
1 1
20 15

1 1
10 5
*
3
♦
1
5

1 1
10 15


1 1
20 25
Fig. 815
Since the period is 10, that portion of the graph in — 5 < x < 5 (indicated heavy in
Fig. 815 above) is extended periodically outside this range (indicated dashed). Note that f(x)
is not defined at x = 0, 5, —5, 10, —10, 15, —15, etc. These values are the discontinuities of f(x).
(b) f{x)
sin x ^ x ^ nr
77 < x < 2tt
Period = 2tt
/<*)
• \
/ \
/ . \. . . 
. J
s
\
/ »
s
/
j
St
—2w —r
V
2r 3a
Av
Fig. 816
Refer to Fig. 816 above. Note that f(x) is defined for all x and is continuous everywhere.
8.13. Prove 1 sin — y
s
dx
■ r
cos — j dx
. kirX j
sin —j— ax
kirX ,
cos — j— dx
I kirX
■j cos—j
kir I
I . kvx
kir I
= if fc= 1,2,3,
j — cos kir + j— cos (— kir)
Kir kit
=
7— sin kir — t~ sin (— kir)
kir kir
=
8.14. Prove
. r l mn
(a) 1 cos — j
(b) f ' si
mrrX fltrX , C
cos—j—dx — \
. mrrX . nirx ,
sin — i— sin — i— dx
m¥= n
1 m = n
mrrX TlirX , n
sin — 5— cos —j dx =
where m and n can assume any of the values 1, 2, 3, ... .
(a) From trigonometry: cos A cos B = £{cos (A  B) + cos (A + B)}, sin A sin B = ^{cos (A  B) 
cos (A + B)}.
Then, if m ¥= n, by Problem 8.13,
Similarly if m ¥= n,
CHAP. 8] APPLICATIONS TO VIBRATING SYSTEMS, ROCKETS AND COLLISIONS 205
If m = n, we have
j^ COS— COS — dx = J^l+COS—jd* = I
J. mirx . nwx , 1 C /.. 2mrx\ .
sm r sm r dx = gj (lcos— )d* = I
i • *  ■'i \ * /
Note that if m = « = these integrals are equal to 2/ and respectively.
(6) We have sin A cosB = ^{sin(AB) + sin(A+B)}. Then by Problem 8.13, if m + n,
C . mirx nirx , If J . (m — n)vx , . (m + n)irx l ,
J_ t sin _~ cos r dx = 2 J_ t \ sin — r— + *™ — i 1 )** = o
If m = n, rl i
( o,~ m ^ x „~„ nvX j 1 I 2mrx ,
I sin — =— cos y dx = k I sm — i — do; =
The results of parts (a) and (6) remain valid even when the limits of integration — I, I are
replaced by y, y + 21 respectively.
8.15. If «, , N
f[x) = A + 2, ( «nCos— y h b n sm— J— J
prove that by making suitable assumptions concerning term by term integration of
infinite series, that for n = 1, 2, 3, . . . ,
(a) a n = \^ f{x)co^dx f (b) b n = \ f f(x)sin^dx, (c) A = ^.
(a) Multiplying
f( x) = A + J^^cos^f + 6 n sin^ (i)
by cos  and integrating from Z to I, using Problem 8.14, we have
J%(*)coB*p«fe = A ^ cos^f da (f)
= a m l if m¥=
Thus
~ 7 J '^ cos— — da; if m = 1, 2, 3, . . .
(6) Multiplying (1) by sin^p and integrating from I to I, using Problem 8.14, we have
£/(«). in Spcfc = A J' sin^fd* (Jf)
+ Si {«*£ sin^cos^fd, + 6„£ sin^sin^f,
— 6 m Z
206 APPLICATIONS TO VIBRATING SYSTEMS, ROCKETS AND COLLISIONS [CHAP. 8
Thus b m = j I f(x) sin— j^dx if m = 1,2, 3, ...
(c) Integration of (1) from — I to Z, using Problem 8.13, gives
f /(«)<*» = 2AI or A = ^J f(x) dx
If 1 _ «o
Putting m = in the result of part (a), we find o = t f(x) dx and so A — ~ .
The above results also hold when the integration limits —I, I are replaced by y, y + 21.
Note that in all parts above we have assumed interchange of summation and integration.
Even when this assumption is not warranted, the coefficients a m and b m as obtained above are
called Fourier coefficients corresponding to f(x), and the corresponding series with these values
of a m and b m is called the Fourier series corresponding to f(x). An important problem in this
case is to investigate conditions under which this series actually converges to f(x). Sufficient
conditions for this convergence are the Dirichlet conditions given on page 197.
8.16. (a) Find the Fourier coefficients corresponding to the function
TO 5<z<0
f(x) = i Period = 10
n ' [3 0<z<5
(b) Write the corresponding Fourier series.
(c) How should f(x) be defined at x = 5, x = and x = 5 in order that the Fourier
series will converge to f(x) for 5 ^ x ^ 5?
The graph of f(x) is shown in Fig. 817 below.
I
— Period ■
10
— r
10
Fig. 817
(a) Period = 21  10 and I  5. Choose the interval y to y + 21 as 5 to 5, so that y = —5. Then
a n = \f + 2l f(x)cos^dx = J 5 /(*)cos^
y _5
=  { f (0) cos^f dx + Jf * (3) cos^f «fa =  jT cos^ dx
= f^sin^f = if n#0
5 V n^r o
o
'5 n .. o /5
3 r 5 ovx , 3 r 5 , _
If n = 0, a n = a = r 1 cos = da; = g I da; = 3.
b n  yj /(*) smy das = gj /(as) sin g das
7 —5
3 / 5 mrx\ I 5 _ 3(1  cos rnr)
I cos > l —
5 \ nir 5 /
CHAP. 8] APPLICATIONS TO VIBRATING SYSTEMS, ROCKETS AND COLLISIONS 207
(b) The corresponding Fourier series is
a ° i v< ( ~ nirX i i. • nirx\ 3 , ~ 3(1 — cos nv) . nwx
 + ^ («n cos — + b n sm— j = 2 + B 2 ^ smg
3 , 6 / . irx . 1 . 3tr« . 1 . 5vx ,
= 2 + ^( Sin 5 + 3 Sin 5 + 5 Sm T + '
(c) Since f(x) satisfies the Dirichlet conditions, we can say that the series converges to f(x) at all
points of continuity and to " x + °^ " x ~ °^ at points of discontinuity. At * = 5, and 5,
which are points of discontinuity, the series converges to (3 + 0)/2 = 3/2 as seen from
the graph. If we redefine f(x) as follows,
'3/2 x = 5
5 < x <
f(x) = < 3/2 * = Period = 10
3 < x < 5
.3/2 x = 5
then the series will converge to f(x) for — 5 ^ x ^ 5.
8.17. If f(x) is even, show that (a) a n =  J" f(x) cos^dx, (b) b n = 0.
(a) «» = J f /(«) cos^ cfe  f J"° /(») cos ^ ^ + I J' /(*) cos^ dx
Letting x = —u,
f /"/<*) cos**<fc = ljV*)cos(^)d % = lJ*V)cos^fd M
since by definition of an even function f(—u) = f(u). Then
«* = y//(«)coB^d« +J , /(*)eos^«te  fff(x)cos^dx
W *n = 7 £ /(^)sin^f d* = J° /(*)sin^<*s + \f f(x)sm?fLdx (1)
If we make the transformation x = u in the first integral on the right of (I), we obtain
\ J° fix) sin 2p ifa =  J* f{u) sin (=f^) <fc* = ] f f( u ) sin ^ <* M
= j f f(u)Bin^du = \ f l f(x)sin^dx (2)
where we have used the fact that for an even function f(u) = f(u) and in the last step that
the dummy variable of integration u can be replaced by any other symbol, in particular x. Thus
from (1), using (2), we have
1 j , M . nnx . 1 f l ,. . . mtx ,
~lj f( x > sm —j— dx + l I /(*) sin— zdx =
8.18. Expand f(x)  x, < x < 2, in a half range (a) sine series, (b) cosine series.
(a) Extend the definition of the given function to that of the odd function of period 4 shown in
Pig. 818 below. This is sometimes called the odd extension of /(»). Then 21 — 4, 1 = 2.
208
APPLICATIONS TO VIBRATING SYSTEMS, ROCKETS AND COLLISIONS [CHAP. 8
/
/
+
/
/
/
/
/(*)
O
1 71 1 7
6 /4 2 y
/ '
/
/
/
/
if
/
/
/
/
/
/
/
/
Fig. 818
Thus a n ~ and
b
n = 1 J /(») sin —j— dx
r
. nirx .
. x smy dx
~0
{w(^¥)«(k
—4 , nirx \
—5 sin g )
cos«jt
Then
,, v v< — 4 . %7ra;
/(a;) — 2i — cos nir sin — — —
n = 1 %7T 2
4 / . 773 1 . 2ttx , 1 . 3jra;
 sin T 2 S in^+ gSin^
(b) Extend the definition of f(x) to that of the even function of period 4 shown in Fig. 819 below.
This is the even extension of f(x). Then 21 = 4, 1 = 2.
Fig. 819
Thus b n = 0,
,. N nirx , 2 ("
a; cos— y dx
. . / 2 . rura \ M v / —4
(*) — smj ) — (1) 5
\71tt 2 / \n 2 ir
—4 »Hr£C
r 2
If n = 0, a = I
•M/V
asda; = 2.
r (cos 1177 — 1) if n ¥^
Then
/(a?) = 1 + > ^^(cosrair — 1) cos— r—
5irx
wx , 1 SwX 1 „„„,
2V COS T + 32 COS "g + g2COS^~ +
It should be noted that the given function f(x) = x, < x < 2, is represented equally well
by the two different series in (a) and (6).
CHAP. 8] APPLICATIONS TO VIBRATING SYSTEMS, ROCKETS AND COLLISIONS 209
SOLUTIONS OF VIBRATING STRING PROBLEMS
8.19. Find the transverse displacement of a vibrating string of length I with fixed endpoints
if the string is initially given a displacement f(x) from its equilibrium position and
then released.
Let the transverse displacement of any point x of the string at time t be Y(x, t). Since the
ends x — and x — I of the string are fixed, we must have Y(0, t) = and Y(l, t) = 0. Since
the initial displacement is f(x), we have Y(x, 0) = f(x); and since the initial velocity is zero, we
have Y t (x, 0) = where Y t denotes the partial derivative with respect to t. We must thus solve
the boundaryvalue problem
&X.  o d2Y
a<2 ~ c dx* w
Y(0,t) = 0, Y(l,t) = 0, Y(x,0)=f(x), Y t (x,0) = (2)
Assume a solution to (1) of the form Y = XT where X depends only on x and T depends
only on t. Then substituting into (1), using X" to denote d 2 X/dx 2 and T" to denote d 2 T/dt 2 , we have
X T" = c 2 X" T
X" _ T"
X ~ c 2 T
(3)
Since one side depends only on x and the other side depends on t while x and t are independent,
the only way in which (2) can be valid is if each side is a constant, which we shall take as
X 2 . Thus
*?  II  _ x2
X ~ c 2 T ~
or X" + \*X = 0, T" + \ z cH =
These equations have solutions
X = A x cos \x + Bi sin Xx, T = A 2 cos Xct + B 2 sin Xct
Thus a solution is given by
Y(x, t) = XT = {A x cos \x + B x sin \x)(A 2 cos Xct + B 2 sin Xct) (i)
From the first condition in (2), we have
B X {A 2 cos Xct + B 2 sin Xct) —
so that A t = [since if the second factor is zero then the solution is identically zero, which we
do not want]. Thus
Y(x, t) = B x sin Xx (A 2 cos Xct + B 2 sin Xct)
= sin Xx (6 cos Xct + a sin Xct) (5)
on writing B t A 2 — b, B X B 2 — a.
Using the second condition of (2 ) in (5), we see that sin \l = or Xl = nir where
n = 1,2,3, ... . Thus X = mr/l and the solution so far is
v(„ t \ — „• nwx ( , nirct . . nvct\ , ,
x (x, t) — sin —j I b cos — j — + a sin — j— j (6)
By differentiating with respect to t, this becomes
v /„ *\ _ „•», nirX ( nircb . nvct nirca nirct\
Y t {x,t)  sin—^ TsnT +  r cos r J
so that the fourth condition in (2) gives
Y t (x,0) = S in^(™p) =
from which a = 0. Thus (6) becomes
v/~ *\ — u • n7rX nnct
Y(x, t) — b sin j cos — y— (7)
210 APPLICATIONS TO VIBRATING SYSTEMS, ROCKETS AND COLLISIONS [CHAP. 8
To satisfy the third condition of (2) we use the fact that solutions of (7) multiplied by constants
as well as sums of solutions are also solutions [the superposition theorem or principle for linear
differential equations]. Thus we arrive at the solution
T7/ ja v u nvx nvct ,„.
Y(x, t) = 2i b n sin —j— cos — — (8)
n=l I I
Using the third condition of {2) in (8) we must have
00
Y(x,0) = f(x) = 2 6„ sin^ (9)
n=l I
But this is simply the expansion of f(x) in a Fourier sine series and the coefficients are given by
b n = T I f( x ) sin— ^— dx
1 Jo l
Thus the solution is given by
Y(x,t) = ^1 jj* l /(x)sin^^jsin^cos^ (10)
The method of solution assuming Y — XT is often called the method of separation of variables.
8.20. A string with fixed ends is picked up at its
center a distance H from the equilibrium
position and released. Find the displace
ment at any position at any time.
From Fig. 820 we see that the initial displace
ment of the string is given by
(2Hx/l 0<x^l/2
Y(x,0) = f(x) = \ 2H{l . x)/l /2 Si * ^ J
NOW n fl
b n = j I f(x) sin j— dx
f f l/2 2Hx . nirx , , C X 2H . mrx .
< I — j— sm j— dx + I —j (t — sc) sm —r~ dx
 %K sin ( w "/ 2 )
v 2 n 2
on using integration by parts to evaluate the integrals. Using this in equation ilO) of Problem 8.19,
we find
„. JX SH ^ sin imr/2) . nvx nirct
Y{x, t) = —z 2i 9 sm ~T~ cos — r~
* 2 n=l n l l
SH f 1 . nX vet 1 . 3rrx Sirct 1 . 5naj 5vct
= ^2\Y 2Sm T c0S r ~ ¥ sm ~r cos ~r + 52 sin_ r cos ~r "
8.21. Find the normal frequencies and normal modes for the vibrating string in
Problem 8.20.
The normal mode corresponding to the lowest normal frequency is given by the first term in
the solution of Problem 8.20, i.e.,
8H . »■»_._. vet
The frequency is given by f x where
sm —= cos ,
r 4 l <>
2Wi = T or A = « ~ «\7
CHAP. 8] APPLICATIONS TO VIBRATING SYSTEMS, ROCKETS AND COLLISIONS
211
Since the cosine varies between —1 and +1, the mode is such that the string oscillates as in
Fig. 821 from the heavy curve to the dashed curve and back again.
Fig. 821
Fig. 822
The next higher frequency is given by the mode corresponding to the next term in the
series which, except for sign, is
81? . 3irx Zirct
^2 sin— cosy
In this case the frequency is given by
2W 3 _ — or fs = Yl = 2^
The mode is indicated in Fig. 822.
The higher normal frequencies are given by
f * = 2z\7' fl = 2i\7> '"
The amplitudes of modes corresponding to the even frequencies
/a = 2!\a' /4 = 21 \ 7' *"
are zero, so that these frequencies are not present. In a general displacement, however, they
would be present.
Because of the fact that all higher normal frequencies are integer multiples of the lowest
normal frequency, often called the fundamental frequency, the vibrating string emits a musical
note. The higher frequencies are sometimes called overtones.
8.22. Find the transverse displacement of a vibrating string of length I with fixed end
points if the string is initially in the equilibrium position and is given a velocity
distribution defined by g(x).
In this case we must solve the boundaryvalue problem
a 2 r _ 2 <py
3<2 ~ C 5*2
CO
Y(0,t) = 0, Y(l,t)=zO, Y(x,0) = 0, Y t (x,0) = g(x)
The method of separation of variables and application of the first two conditions of (2\ vields
as in Problem 8.19, ;
V(rr +\ — „J„ nirX ( U niTCt i • % "" C * \
X \x, t) — sin —j— 6 cos — : — + a sin — =— )
I \" wo I "■■ " ai " i J
However, in this case if we apply the third condition of (2) we find 6 = 0, so that
w~ *\ _ • nvX . nvct
Y(x, t) = a sin — j— sin ^~
To satisfy the fourth condition we first note that the superposition principle applies, so that we
arrive at the solution
x{x, t) — 2, a n sin —f— sin
»=i '
I
212 APPLICATIONS TO VIBRATING SYSTEMS, ROCKETS AND COLLISIONS [CHAP. 8
From this we have by differentiation with respect to t,
v , . N _ £, nirca n _ nwX n ct
Y t (x, t) — 2i — i — sm ~t~ cos ~ ; —
n=X l I I
or Y t (x, 0) = g{x) = 2 — i — sin ~T
w=l * l
Then by the method of Fourier series we see that
mrca n 2
I
Thus the required solution is, on using (4) in
2 C , \ • n T?% j 2 C . nirx .
7 I g(x) sm—7— dx or a n ■— I g(x) sm—j— ax
I J I nirc J I
<*)
sin — . — (o)
v/ i\ S I 2 f , . . mrx . 1 . nvx .
Y(x,t) = 2 x — J o ir(«) ain r <te sm— si
8.23. Find the transverse displacement of a vibrating string of length I with fixed end
points if the string initially has a displacement from the equilibrium position given
by f(x) and velocity distribution given by g(x).
The solution to the given problem is the sum of the solutions to the Problems 8.19 and 8.22.
Thus the required solution is
nirx nirct
sm — — cos
Y(x,t) ■= 2 "I7 j /(*) sin^j^ dx
I WB I
nirX . nrrct
+ 2 1 I g(x)sm^dx> sin— r sm
n=l «J/ W I I
MISCELLANEOUS PROBLEMS
8.24. A particle is dropped vertically on to a fixed horizontal plane. If it hits the plane
with velocity v, show that it will rebound with velocity — cv.
The solution to this problem can be obtained from the results of Problem 8.6 by letting m 2
become infinite and v 2 = while v x = v [where subscripts 1 and 2 refer to the particle and
plane respectively]. Then the respective velocities after impact are given by
{{mjm^ — e}v
lim v( = lim ., , , , — r— = — ev
m 2 +°o m 2 +oo 1 + {m 1 /m 2 )
(7n 1 /m 2 )(l + e)v
lim Vo = lim — ; — r— : 1 — r =
Thus the velocity of the particle after impact is ev. The velocity of the plane of course
remains zero.
8.25. Suppose that the particle of Problem 8.24 is dropped from rest at a height H above
the plane. Prove that the total theoretical distance traveled by the particle before
coming to rest is given by #(1 + e 2 )/(l  e 2 ).
Let v be the speed of the particle just before it hits the plane. Then by the conservation of
energy, \mv 2 + = + mgH or v 2  2gH. Thus by Problem 8.24 the particle rebounds with
speed ev and reaches a height {ev) 2 /2g = e 2 H. It then travels back to the plane through the
distance e 2 H. Thus on the first rebound it travels through the distance 2e 2 H.
By similar reasoning we find that on the second, third, . . . rebounds it travels through the
distance 2e 4 H, 2e 6 H , .... Then the total theoretical distance traveled before coming to rest is
H + 2e 2 H +2e*H + 2e6#+ ••• = H + 2«*ff(l + e 2 + e* + • • •) = H + ^^ = #(fz"£)
using the result l + r + r 2 + r s + ~ 1/(1  r) if \r\ < 1.
CHAP. 8] APPLICATIONS TO VIBRATING SYSTEMS, ROCKETS AND COLLISIONS
213
8.26. Two particles having masses m and M
are traveling on the x axis (assumed
frictionless) with velocities vd and V\i
respectively. Suppose that they collide
and that after the collision (impact) their
velocities are v 2 i and V 2 i respectively.
Prove that the velocities of the center of
mass before and after collision are equal.
By the conservation of momentum,
Total momentum before impact
mt^i + MVxi
or mVl + MV t
O
m
Fig. 823
total momentum after impact
mv 2 \ + MV 2 i
mv 2 + MV 2
Let x and X be the respective coordinates of the particles. Then the center of mass is given
by f = (mx + MX)/(m + M).
The velocity of the center of mass before impact is
r x — (mv, + MVJ/im + M)
The velocity of the center of mass after impact is f 2 = (mv 2 + MV 2 )/(m + M). Thus ^ = *f 2 .
8.27.
A particle of mass m slides down a frictionless incline of angle <*, mass M and
length L which is on a horizontal frictionless plane [see Fig. 824]. If the particle
starts initially from rest at the top of the incline, prove that the time for the particle
to reach the bottom is given by
J 2L(M + m si n 2 a)
(M + m)g sin a
Choose a fixed vertical xy coordinate
system as represented in Fig. 824. Let R be
the position vector of the center of mass C
of the incline, A the (constant) vector from
C to the top of the incline, and s the position
vector of the particle relative to the top of
the incline. Then the position vector of par
ticle m with respect to the fixed coordinate
system is R + A + s. Since the only force
acting on the particle is the weight mg of the
particle, we have by Newton's second law
applied to the particle,
d 2 ,„ . . . .
mg (1)
mjp (R+A + s) =
dt* + dt 2
(2)
Fig. 824
Writing R Xi+Yj, g  ^j and s = ss lt where Sl is a unit vector down the incline in the
direction of s, (2) becomes
d 2 X . d%
dt 2 * + dt 2 Sl ~
Multiplying by s, • , this becomes
d 2 X cPs
dt 2 Sl ' * + dt 2 Sl ' Sl
91
ffSi ' J
d 2 X d 2 s
— COSa +__
9 sin a
(»)
214 APPLICATIONS TO VIBRATING SYSTEMS, ROCKETS AND COLLISIONS [CHAP. 8
Since the net horizontal force acting on the system consisting of the particle and incline is zero, the
total momentum in the horizontal direction before and after the particle starts sliding is zero. Then
M^i + m~CR + A + s)'i =
at dt
This can be written as
(M + m) j m j cos a = (4)
Differentiating (4) with respect to * and solving for d 2 X/dt 2 , we find
d 2 X _ m cos a cPs
dt 2 ~ M + m dt 2
Substituting into (3) yields
(j2 s (M + m)g sin a (M + m)g sin a
dt 2 ~~ M + m — m cos 2 a ~ M + m sin 2 a
Integrating (6) subject to the conditions 8 = 0, ds/dt = at * = 0, we find
1 C(M + m)g sin a]
(5)
(6)
s =
t 2
2 J I + m sin 2 a
which, when 8 — L, yields the required time.
8.28. Solve the vibrating string Problem 8.19 if gravity is taken into account.
The boundaryvalue problem is
d 2 Y 2 d 2 Y m
F(0, t) = 0, Y(l, t) = 0, Y(x, 0) = f(x), Y t (x, 0) = (2)
Because of the term — g the method of separation of variables does not work in this case. In
order to remove this term, we let
Y(x,t) = Z(x,t) + f(x) (3)
in the equation and conditions. Thus we find
d 2 Z „d 2 Z , „ „ m
Z(0, t) + \fr(0) = 0, Z(l, t) + f(l) = 0, Z(x, 0) + f(x) = /(*), Z t (x, 0) = (5)
The equation (4) and conditions (5) become similar to problems already discussed if we choose $
such that
efy"  g = 0, *(0) = 0, f(l) = (6)
In this case (4) and (5) become
dt 2 c dx 2 Kn
Z(0, t) = 0, Z(l, t) = 0, Z(x, 0) = f{x)  f(x), Z t (x, 0) = (8)
Now from (6) we find f"  g/c 2 or $(x) = gx 2 /2c 2 + c x x + c 2 ; and since ^(0) = 0, f(l) = 0,
we obtain c 2 = 0, c± = —gl/2c 2 . Thus
CHAP. 8] APPLICATIONS TO VIBRATING SYSTEMS, ROCKETS AND COLLISIONS 215
The solution to equation (7) with conditions (8) is, as in Problem 8.19,
Z(x, t) = 2 i  f [/(*)  *(*)} sin ^ dx
. nirx nirct
sin — — cos
I I
and thus
Y(x, t) = ^2  f [/<*)  JL ( X 2 _ fa) J sin « da A sin 2^ cos «** + J?L (jB 2 _ lx)
8.29. Assume that a continuous string, which is fixed at its endpoints and vibrates
transversely, is replaced by N particles of mass m at equal distances from each other.
Determine the equations of motion of the particles.
We assume that the particles are con
nected to each other by taut, elastic strings
having constant tension T [see Fig. 825].
We also assume that the horizontal distances
between particles [i.e. in the direction of the
unit vector i] are equal to a and that the
transverse displacement [i.e. in the direction
of the unit vector j] of particle v is Y v . We
assume that there is no displacement of any
particle in direction i or — i. Fig. 825
Let us isolate the rfh particle. The forces acting on this particle are those due to the (v  l)st
and {v + l)st particles. We have
fY — Y _ \
Transverse force due to {v  l)st particle = —T ( — — ) j
fY — Y \
Transverse force due to {v + l)st particle = —T ( — — ) j
Then by Newton's second law the total transverse force acting on particle v
Si = *(*^=0ir(*^)i
m d*
d?Y v T
m W = (Y*i2Y v + Y v + 1 )
le  Y» = £(^i2^+^ + i) (1)
To take into account the fact that the endpoints are fixed, we assume two particles cor
responding to v = and r = N+l for which Y = 0, Y N + 1 = 0. Then on putting „ = 1 and
v = N in equation (1), we find
^ = ^ 2Y i + Y ^> ? s = ^(Y N _ 1 2Y N ) W
8.30. Obtain the secular determinant condition for the normal frequencies of the system
of particles in Problem 8.29.
Let Y v = A v cosat in equations (1) and (2) of Problem 8.29. Then after simplifying we find
A„_! + (2 ma a 2 /T)A v  A v + 1 = v = 2, . . ., N 1 (l)
(2  ma«?IT)A x  A 2 = 0, ~A N _ 1 + (2  ma^/T)A N = (2)
Putting 2  mtuP/T = c (3)
these equations can be written
cA 1 A 2 = 0, A 1 + cA 2 A 3 = 0, ..., A N _ 1 + cA N =
216
APPLICATIONS TO VIBRATING SYSTEMS, ROCKETS AND COLLISIONS [CHAP. 8
Then if we wish solutions such that A v ¥= 0, we must require that the 2Vth order determinant
of the coefficients be zero, i.e.,
A W =
c
1
..
1
c
1
..
1
c
1
..
..
. 1
c
1
..
1
c
=
The normal frequencies are obtained by solving this equation for the N values of w 2 .
Although we have used Y v = A v cos cot, we could just as well have assumed Y v = B v sin wt
or Y v = A v cos at + B v sin at or Y v = C v e iat . The secular determinant would have come out to be
the same [compare the remarks at the end of Problem 8.2(6)].
8.31. Prove that the normal frequencies in Problem 8.30 are given by
2T
<D„ —
ma
cos
N + l
a= 1, ...,N
By expanding the determinant A N of Problem 8.30 in terms of the elements in the first row,
we have
A N = cA N i  A N _ 2 (I)
Also,
A x = c, A 2 = c 2  1
Putting N = 2 in (i), we see that equations (2) are formally satisfied if we take A = 1. Thus
conditions consistent with (1) and (2) are
A = 1, A! = c (3)
To solve the difference equation (1), assume that A N = p N where p is a constant to be
determined. Substituting this into (1), we find on dividing by p N ~ 2 ,
p 2 — cp + 1
If we call c = 2 cos 0, then
V =
c ± Vc 2 4
,±i0
p = cos 6 ± t sin e
Thus solutions of the difference equation are
( e ifl)N = e iN0 = cos Ne + i sin Nff and (e~ ie ) N = e~ Nie = cos No  i sin No
Since constants multiplying these solutions and sums of solutions are also solutions [as in
the case of linear differential equations], we see that the general solution is
A N = G cos Ne + H sin Ne (4)
Now from equations (3) we have A = 1, A t = 2 cos e so that G = 1, H = cot e. Thus
sin Ne cos e sin (N + l)e
= cos Ne +
sins
sine
This is equal to zero when sin (N + 1)6 = or e = aw/(N +1), a = l,...,N. Thus using
of Problem 8.30, we find
2 2T fl «*
o 2 = 1 — COS t. 7 i .,
« ma\ N + l
(5)
(6)
CHAP. 8] APPLICATIONS TO VIBRATING SYSTEMS, ROCKETS AND COLLISIONS 217
8.32. Solve for A v in Problem 8.30 and thus find the transverse displacement Y v of particle v.
From equations (1) and (2) of Problem 8.30 we have [on using the normal frequencies (6) of
Problem 8.31 and the superscript a to indicate that the A's depend on a],
Ail\ + 2AJ"> cos^^  A%\ = (1)
together with the end conditions
K°° = 0, A^l, = (2)
The equation (1) subject to conditions (2) can be solved in a manner exactly like that of Problem 8.31,
and we find
I (a)  ■ av *
A^ = C«sin
N + l
where C a are arbitrary constants. In a similar manner if we had assumed Y v = B v sin ut [see
remarks at the end of Problem 8.30] we would have obtained
avir
>(«) _
D„ sin
N+l
Thus solutions are given by
C a sin — cos u>t and D a sin — sin ut
N+l N+l
and since sums of solutions are also solutions, we have
N
*v — 2i sin \C a cos oxt + D a sin «t)
a = l iV + 1
The constants C a and D a are determined from initial conditions.
The analogy with the continuous vibrating string is easily seen.
Supplementary Problems
VIBRATING SYSTEMS OF PARTICLES
8.33. Find the normal frequencies of the vibrations in Problem 8.1, page 197, if the spring constants
and masses are all different.
8.34. Two equal masses m on a horizontal frictionless
table as shown in Fig. 826 are connected by equal
springs. The end of one spring is fixed at A
and the masses are set into motion, (a) Set up the ^JL~^ m * m
equations of motion of the system. (6) Find the A \ r Q'O'OQ^ Q ' QOOQ ^ Q
normal frequencies of vibration, (c) Describe the
normal modes of vibration.
a». (» A = 3^± JI /, = 4±1 JI Fi * 8  26
Air \ m ' z 4v y m
8.35. Work Problem 8.34 if the spring constants and masses are different.
8.36 Two equal masses m are attached to the ends of a spring of constant k which is on a horizontal
frictionless table. If the masses are pulled apart and then released, prove that they will vibrate
with respect to each other with period 2iryfm/2tc.
8.37. Work Problem 8.36 if the masses are different and equal to M t and M 2 respectively.
Ans. 2irV^/K where n = M 1 M 7 J{M l + M 2 )
218
APPLICATIONS TO VIBRATING SYSTEMS, ROCKETS AND COLLISIONS [CHAP. 8
8.38. In Fig. 827 equal masses m lying on a horizontal
frictionless table are connected to each other and to
fixed points A and B by means of elastic strings of
constant tension T and length I. If the displacements
from the equilibrium position AB of the masses are
Y l and F 2 respectively, show that the equations of
motion are given by
Y, = k(Y 2 2Y 1 ), Y 2 = k(Y 1 ~2Y 2 )
where k — ST I ml.
Fig. 827
.39. Prove that the natural frequencies of the vibration in Problem 8.38 are given
respectively by
JL /3T
2w \ ml
and describe the modes of vibration.
and
1 9T
2tt V ml
.40. Find the normal frequencies and normal modes of vibration for the system of
particles of masses m x and m 2 connected by springs as indicated in Fig. 828.
A
CHANGING MASS. ROCKETS
8.41. (a) Prove that the total distance traveled by the rocket of Problem 8.5 in time t is given by
v <t +
m
at\ /m
In
V m o
 gt 2
2 y
(b) What is the maximum height which the rocket can reach and how long will it take to
achieve this maximum height?
8.42. Suppose that a rocket which starts from rest falls in a constant gravitational field. At the
instant it starts to fall it ejects gas at the constant rate a in the direction of the gravitational
field and at speed v with respect to the rocket. Find its speed after any time t.
m ^
Ans. gt — v In
at J
8.43. How far does the rocket of Problem 8.42 travel in time tl
Ans. — gt 2
m — at
t + ( — ) In
m — at
m
8.44. Describe how Problem 8.42 can be useful in making a "soft landing" on a planet or satellite?
8.45. Discuss the motion of a twostage rocket, i.e. one in which one part falls off and the other
rocket takes over.
COLLISIONS OF PARTICLES
8.46. A gun fires a bullet of mass m with horizontal velocity v into a block of wood of mass M which
rests on a horizontal frictionless plane. If the bullet becomes embedded in the wood, (a) determine
the subsequent velocity of the system and (b) find the loss in kinetic energy.
Ans. (a) mv/(M + m) (b) mMv 2 /2(M + m)
8.47. Work Problem 8.46 if the block is moving away from the gun with velocity V.
8.48. A ball which is dropped from a height H onto a floor rebounds to a height h < H. Determine
the coefficient of restitution. Ans. Jh/H
8.49. A mass m 1 traveling with speed v on a horizontal plane hits another mass m 2 which is at rest.
If the coefficient of restitution is e , prove that there is a loss of kinetic energy equal to
m 1 m 2 (l — e 2 )v 2 /2(m 1 + m 2 ).
CHAP. 8] APPLICATIONS TO VIBRATING SYSTEMS, ROCKETS AND COLLISIONS 219
8.50. A billiard ball strikes another billiard ball obliquely at an angle of 45° with their line of
centers at the time of impact. If the coefficient of restitution is 1/2, find the angle at which
the first ball will "bounce off". Ans. tan" 1 (3/5)
8.51. Let the masses of two colliding particles be m 1 ,m 2 and their respective velocities before impact
be v t ,v 2 . If the coefficient of restitution is e, prove that the loss in kinetic energy as a result
of the collision is •= , (v x  v 2 ) 2 (l  c 2 ).
8.52. Prove that the momentum which is transferred from the first particle of Problem 8.51 to the second
is _, (1 + e )(v 1 — v 2 ).
8.53. A ball is dropped from a height h above a horizontal plane on to an inclined plane of angle a which
is resting on the horizontal plane. Prove that if the coefficient of restitution is e, then the ball will
next hit the incline at a point which is at a distance 4e(l + e)h sin « below the original point
of impact.
FOURIER SERIES, ODD AND EVEN FUNCTIONS, FOURIER SINE AND COSINE SERIES
8.54. Graph each of the following functions and find their corresponding Fourier series using properties
of even and odd functions wherever applicable.
[ 8 < x < 2
(a) f(x) = j_ 8 2 < x < 4 Period 4 (") fW ~ 4x ' ° < x < 10 > Period 10
IX Cx A^x^O (2x 0^x<3
{b) f{X) = { x ^ x £ 4 Peri ° d 8 W m =  _ s<x<0 Period 6
Ans . (o) W  (1 cos n,) s . n n^ _ 40  1 «
(h\ 2 — V C 1 ~ cos nir) r nvx ... 3 « \ 6(cos nv  1) nirx 6 coanir . nirx\
{b) ** A n* C0S ~T id) 2 + „?> J *P COS "3 nV~ Sin 3
8.55. In each part of Problem 8.54, tell where the discontinuities of f(x) are located and to what value
the series converges at these discontinuities.
Ans. (a) x = 0, ±2, ±4, ... ; (c) x = 0, ±10, ±20, . . . ; 20
(b) no discontinuities (d) x — ±3, ±9, ±15, . . . ; 3
(2 x < a; < 4
8.56. Expand f(x) = < in a Fourier series of period 8.
[16 4<«<8
i 6  J „ E®. j_ 1. <?£» J^ 5 ffa;
3 2 cos 4 ■+■ 52
Ans. JHcos^f + ^cos^F + ^ cos ^ +
8.57. (a) Expand f(x) = cos x, < x < v, in a Fourier sine series.
(6) How should f(x) be defined at x = and a; = v so that the series will converge to fix) for
=5 £C ^ 7T?
a 1 \ 8 v ttsin2/i# / ,, ,,^ ,, ^
Ans. (a)  n 2 4n21 (6) /(0) = fW) =
8.58. (a) Expand in a Fourier series f(x)  cos a;, < x < v if the period is w, and (6) compare with
the result of Problem 8.57, explaining the similarities and differences if any.
Ans. Answer is the same as in Problem 8.57.
220 APPLICATIONS TO VIBRATING SYSTEMS, ROCKETS AND COLLISIONS [CHAP. 8
fx < x < 4
8.59. Expand f{x) — < n J in a series of (a) sines, (6) cosines.
[8 — x 4 < x < 8
. , . 32 £, 1 , «ir . nvx /L , 16 « / 2 cos nir/2 — cos mr — l\ mrx
Ans. (a) v% 2 x ^ Bm T sin— (6) ^2 (  t ) cos —
THE VIBRATING STRING
8.60. (a) Solve the boundaryvalue problem
d 2 Y d 2 Y
1W = 4 M 0<x<,,t>0
Y x (0,t) = 0, Y(r,t) = h, Y(x,0) = 0, Y t (x,0) =
(6) Give a physical interpretation of the problem in (a).
Aws. F(a, *) = — 2 „ ' sin (w  4)« sin (2n  l)t
t „=i 2w  1 2
8.61. Solve the boundaryvalue problem
Y tt = Y xx  g < x < *, t >
F(0, t) = 0, Y{ir,t)  0, Y(x,0) = px(irx), Y t (x,0) =
and interpret physically.
Ans. Y(x, t) = 4(2/ta? + flr) % /o 1 ,^ sin ( 2w ~ !)* cos ( 2w _ !)* ~ i^"' _ x )
ir n =i (2w — 1)^
d 2 Y d 2 Y
8.62. (a) Find a solution of the equation rp = 4 ^j which satisfies the conditions F(0, t) = 0,
Y( v , t)  0, Y(x, 0) = 0.1 sin x + 0.01 sin 4x, Y t (x, 0) = for < x < ir, t > 0. (6) Interpret physi
cally the boundary conditions in (a) and the solution.
Ans. (a) Y(x, t) = 0.1 sin x cos 2t + 0.01 sin 4a; cos St
Q2V d 2 Y
8.63. (a) Solve the boundaryvalue problem ^ = 9 ^ subject to the conditions Y(0, t) = 0,
F(2, t) = 0, Y(x, 0) = 0.05x(2  x), Y t (x, 0) = 0, where < x <2,t> 0. (6) Interpret physically.
/ x v , *v 16 4 1 • (2n  l)irx 3(2nlM
Aws. (a) F(ar, t) = ^j 2 ( 2w _ i)3 sm <j C0S 2
8.64. Solve Problem 8.63 with the boundary conditions for Y(x,0) and Y t (x,0) interchanged, i.e.
Y(x, 0) = 0, Y t (x, 0) = 0.05sc(2  x), and give a physical interpretation.
™ * 3.2 2, 1 . (2w  l)yg . B(2nl)wt
Ans. Y(x, t) =—t 2 x (2^71)4 ««n 2 «* 2
MISCELLANEOUS PROBLEMS
8.65. A spherical raindrop falling in a constant gravitational field grows by absorption of moisture from
its surroundings at a rate which is proportional to the instantaneous surface area. Assuming
that it starts with radius zero, determine its acceleration. Ans. ^g
8.66. A cannon of mass M rests on a horizontal plane having coefficient of friction ft. It fires a
projectile of mass m with muzzle velocity v in a direction making angle a with the horizontal.
Determine how far back the cannon will move due to the recoil.
8.67. A ball is thrown with speed v onto a smooth horizontal plane in a direction making angle a with
the plane. If e is the coefficient of restitution, prove that the velocity of the ball after the impact
is given by v V 1  (1 — e 2 ) sin 2 a in a direction making angle tan 1 (e tan a) with the horizontal.
CHAP. 8] APPLICATIONS TO VIBRATING SYSTEMS, ROCKETS AND COLLISIONS
221
8.68. Prove that the total theoretical time taken for the particle of Problem 8.67 to come to rest is
8.69. Prove that while the particle of Problem 8.27, page 213, moves from the top to the bottom of the
incline, the incline moves a distance {mL cos a)/{M + m).
8.70. Prove that the loss of kinetic energy of the spheres of Problem 8.9 is ^(^i cos 0y — v 2 cos 2 ) 2 (1 ~~ e *)
where fi is the reduced mass m 1 m 2 /{m, 1 + m 2 ).
8.71. Prove that the acceleration of a double incline of mass M [Fig. 829] which is on a smooth
(m 1 sin a x cos ay — m 2 sin o 2 cos a 2 )g
table is given by
M + my sin 2 ay + m 2 sin 2 a 2
8.72. If A is the acceleration of the incline in Problem 8.71, prove that the accelerations of the masses
my(A cos ay + g sin ay) + m 2 (A cos a 2 — g sin a 2 )
relative to the incline are given numerically by ; .
m[ + m 2
Fig. 829
Pig. 830
8.73. A mass in slides down an inclined plane of the same mass which is on a horizontal plane with
coefficient of friction fi. Prove that the inclined plane moves to the right with acceleration equal
to (1  3/t)/(3  /x). See Fig. 830.
8.74. A gun of mass M is located on an incline of angle a which in turn is on a smooth horizontal plane.
The gun fires a bullet of mass m horizontally away from the incline with speed v . Find the
recoil speed of the gun. Ans. (mv cos a)/M up the incline
8.75. How far up the plane will the gun of Problem 8.74 move before it comes to rest if the incline is
(a) frictionless, (b) has coefficient of friction /*?
8.76. A weight W is dropped from a height H above the plate AB of
Fig. 831 which is supported by a spring of constant k. Find the ^
speed with which the weight rebounds.
8.77. A ball is thrown with speed v at angle a with a horizontal plane.
If it rebounds successively from the horizontal plane, determine
its location after n bounces. Assume that the coefficient of restitution
is e and that air resistance is negligible.
Fig. 831
8.78. Work Problem 8.77 if the horizontal plane is replaced by an inclined plane of angle /? and the
ball is (a) thrown downward, (b) thrown upward.
8.79. Obtain the equation (1), page 195, for the vibrating string by considering the equations of motion
for the N particles of Problem 8.29, page 215, and letting N » °°.
8.80. Prove that as N * °° the normal frequencies as given in Problem 8.31, page 216, approach those
for the continuous vibrating string.
8.81. Prove that for ^ x ^ ir,
t 2 f cos 1x
(a) x(ir — x) =
(6) x(v — x) —
6
l 2
cos 4a; cos 6a;
i ~m "i" T^o T
2 2
32
8 /sin x , sin 3a; sin 5a;
l 3
33
5 3
222 APPLICATIONS TO VIBRATING SYSTEMS, ROCKETS AND COLLISIONS [CHAP. 8
8.82. Use Problem 8.81 to show that
n =i w 6 ~ x n 2 12 n =i (2n — l) 3 32
8.83. Prove that Y = f(x + ct) + g(x  ct) is a solution of the equation
d*Y _ „ d 2 Y
dt 2 ° dx 2
and discuss the connection of this solution with the problem of the vibrating string.
T C x /dY\ 2
8.84. (a) Prove that the total potential energy of a vibrating string is V = — I ( r— ) dx.
(b) Thus show that V = ~ 2 n% ( a n cos ~^i •" ^n si n — T  ) •
2
8.85. (a) Prove that the total kinetic energy of the vibrating string is K.E. = ^<r I ( — )
(b) Thus show that K.E. = .. 2 n2 ( a n cos ~i b n sin n J .
(c) Can the kinetic energy be infinite? Explain.
ir 2 T °°
8.86. Prove that the total energy of a vibrating string is E = — tt 2 n2 ( a n + &!»)•
4f n=l
8.87. Find the potential energy, kinetic energy and total energy for the string of (a) Problem 8.20,
page 210, (b) Problem 8.28, page 214.
8.88. If damping proportional to the instantaneous transverse velocity is taken into account in the
d 2 Y dY „ d 2 Y
problem of the vibrating string, prove that its equation of motion is ttj + P ~qT — c z Tg" •
8.89. Prove that the frequencies of vibration for the damped string of Problem 8.88 are given by
y/nW&IP  p 2 /4, n = 1, 2, 3,
8.90. Solve the problem of the damped vibrating string if the string is fixed at the endpoints x = and
x = I and the string is (a) given an initial shape f(x) and then released, (b) in the equilibrium
position and given an initial velocity distribution g(x), (c) given an initial shape f(x) and velocity
distribution g{x).
8.91. Work the problem of the damped vibrating string if gravitation is taken into account.
8.92. Work (a) Problem 8.84(a), (6) Problem 8.85(a), (c) Problem 8.86, (a*) Problem 8.88 for the
case where the string is replaced by N particles as in Problem 8.29, page 215.
8.93. In Fig. 832 the double pendulum system is free to vibrate in a vertical Y///////////
plane. Find the normal frequencies and normal modes assuming small
vibrations.
8.94. Work Problem 8.93 if there is an additional mass m 3 suspended from m 2 by
a string of length Z 3 .
8.95. Generalize the motion of (a) Problem 8.1, (6) Problem 8.34 to N equal
particles and springs.
8.96. In Problem 8.95 investigate the limiting case as N » ». Discuss the
physical significance of the results.
CHAP. 8] APPLICATIONS TO VIBRATING SYSTEMS, ROCKETS AND COLLISIONS 223
8.97. Solve the boundaryvalue problem
d 2 Y _ fl2 F
F(o,t) = o, F(U) = o, r(*,o) = /(«), Y t (*,o) = o
and give a physical interpretation.
8.98. Work Problem 8.97 if the condition Y t {x, 0) = is replaced by Y t (as, 0) = g(x).
8.99. Work Problem 8.97 if the partial differential equation is replaced by
and interpret physically.
d*Y , Q dY 2 d*Y .
■W + Pin = c Jx^ + asmut
8.100. Set up the differential equations and initial conditions for the motion of a rocket in an inverse
square gravitational field. Do you believe these equations can be solved? Explain.
8.101. Two bodies [such as the sun and earth or earth and moon] of masses m x and m 2 move relative
to each other under their mutual inverse square attraction according to the universal law of
gravitation. If r x and r 2 are their position vectors relative to a fixed coordinate system, and
r = r t — r 2 , prove that their equations of motion are given by
.. _ Gm 2 (r 1  r 2 ) .. _ Gm 1 {r 2  r t )
r i ~ ~ r3 ' r 2 ~ ^3
This is called the problem of two bodies.
8.102. In Problem 8.101 choose a new origin at the center of mass of the two bodies, i.e. such that
m 1 r 1 + m 2 r 2 = 0. Thus show that if we let r be the position vector of m x relative to m 2 , then
.. _ G(m x + m 2 )r x .. G^ + m 2 )r 2
r i ~ ^3 ' r 2 ~ ^
,. .. .. G{m 1 + m 2 )r
or, on subtracting, r = ;
r 3
Thus show that the motion of m x relative to m 2 is exactly the same as if the body of mass m 2 were
fixed and its mass increased to m l + m 2 .
8.103. Using Problem 8.102, obtain the orbit of mass m t relative to w 2 and compare with the results of
Chapter 5. Are Kepler's first and second laws modified in any way? Explain.
8.104. If P is the period of revolution of m l about m 2 and a is the semimajor axis of the elliptical path of
m 1 about m 2 , prove that
P2 _ 4tt 2
a 3 G(m l + m 2 )
Compare this result with Kepler's third law: In the case of the earth [or other planet] and sun,
does this modified Kepler law have much effect? Explain.
8.105. Set up equations for describing the motion of 3 bodies under a mutual inverse square law of
attraction.
8.106. Transform the equations obtained in Problem 8.105 so that the positions of the bodies are described
relative to their center of mass. Do you believe these equations can be solved exactly?
8.107. Work Problems 8.105 and 8.106 for N bodies.
Chapter 9 PLANE MO TION
of RIGID BODIES
RIGID BODIES
A system of particles in which the distance between any two particles does not change
regardless of the forces acting is called a rigid body. Since a rigid body is a special case
of a system of particles, all theorems developed in Chapter 7 are also valid for rigid bodies.
TRANSLATIONS AND ROTATIONS
A displacement of a rigid body is a change from one position to another. If during
a displacement all points of the body on some line remain fixed, the displacement is called
a rotation about the line. If during a displacement all points of the rigid body move in
lines parallel to each other the displacement is called a translation.
EULER'S THEOREM. INSTANTANEOUS AXIS OF ROTATION
The following theorem, called Euler's theorem, is fundamental in the motion of rigid
bodies.
Theorem 9.1. A rotation of a rigid body about a fixed point of the body is equivalent
to a rotation about a line which passes through the point.
The line referred to is called the instantaneous axis of rotation.
Rotations can be considered as finite or infinitesimal. Finite rotations cannot be
represented by vectors since the commutative law fails. However, infinitesimal rotations
can be represented by vectors.
GENERAL MOTION OF A RIGID BODY. CHASLE'S THEOREM
In the general motion of a rigid body, no point of the body may be fixed. In such case
the following theorem, called Chasle's theorem, is fundamental.
Theorem 9.2. The general motion of a rigid body can be considered as a translation
plus a rotation about a suitable point which is often taken to be the center of mass.
PLANE MOTION OF A RIGID BODY
The motion of a rigid body is simplified considerably when all points move parallel
to a given fixed plane. In such case two types of motion, called plane motion, are possible.
1. Rotation about a fixed axis. In this case the rigid body rotates about a fixed axis
perpendicular to the fixed plane. The system has only one degree of freedom [see Chap
ter 7, page 165] and thus only one coordinate is required for describing the motion.
224
CHAP. 9]
PLANE MOTION OF RIGID BODIES
225
2. General plane motion. In this case the motion can be considered as a translation
parallel to the given fixed plane plus a rotation about a suitable axis perpendicular to
the plane. This axis is often chosen so as to pass through the center of mass. The num
ber of degrees of freedom for such motion is 3: two coordinates being used to describe
the translation and one to describe the rotation.
The axis referred to is the instantaneous axis and the point where the instantaneous
axis intersects the fixed plane is called the instantaneous center of rotation [see page 229].
We shall consider these two types of plane motion in this chapter. The motion of a
rigid body in three dimensional space is more complicated and will be considered in
Chapter 10.
MOMENT OF INERTIA
A geometric quantity which is of great importance in discussing the motion of rigid
bodies is called the moment of inertia.
The moment of inertia of a particle of mass m about a line or axis AB is defined as
/ = mr 2
where r is the distance from the mass to the line.
The moment of inertia of a system of particles, with
masses mi, ra 2 , . . . , m N about the line or axis AB is defined as
/ = ^m v r 2 = m t r 2 + m 2 r 2 +
+ m N r N
(2)
where r%, r 2 , . . . , r N are their respective distances from AB.
The moment of inertia of a continuous distribution of
mass, such as the solid rigid body % of Fig. 91, is given by
/ = J r 2 dm (S)
where r is the distance of the element of mass dm from AB.
(0
Fig. 91
RADIUS OF GYRATION
N
Let / = ^ m v r 2 be the moment of inertia of a system of particles about AB, and
N v = i
M = ^ m v be the total mass of the system. Then the quantity K such that
2 ™«> r l
K 2 = — =
M
2 m *
(*)
is called the radius of gyration of the system about AB.
For continuous mass distributions (4) is replaced by
K 2 = — =
Jr 2 dm
J dm
(5)
226
PLANE MOTION OF RIGID BODIES
[CHAP. 9
THEOREMS ON MOMENTS OF INERTIA
1. Theorem 9.3: Parallel Axis Theorem. Let / be the moment of inertia of a system
about axis AB and let I c be the moment of inertia of the system about an axis parallel to
AB and passing through the center of mass of the system. Then if b is the distance between
the axes and M is the total mass of the system, we have
/ =
I c + Mb 2
2. Theorem 9.4: Perpendicular Axes Theorem.
xy plane of an xyz coordinate system. Let h, I y and h
the x, y and z axes respectively. Then
h = I X +Iy
Consider a mass distribution in the
denote the moments of inertia about
en
SPECIAL MOMENTS OF INERTIA
The following table shows the moments of inertia of various rigid bodies which arise in
practice. In all cases it is assumed that the body has uniform [i.e. constant] density.
Rigid Body
Moment of Inertia
1. Solid Circular Cylinder
of radius a and mass M
about axis of cylinder.
\Ma?
2. Hollow Circular Cylinder
of radius a and mass M
about axis of cylinder.
Wall thickness is negligible.
Ma*
3. Solid Sphere
of radius a and mass M
about a diameter.
Ma2
4. Hollow Sphere
of radius a and mass M
about a diameter.
Sphere thickness is negligible.
Ma2
5. Rectangular Plate
of sides a and 6 and
mass M about an axis
perpendicular to the plate
through the center of mass.
JLM(a2 + 62)
6. Thin Rod
of length a and mass M
about an axis perpendicular
to the rod through the
center of mass.
^Ma*
COUPLES
A set of two equal and parallel forces which act in
opposite directions but do not have the same line of action
[see Fig. 92] is called a couple. Such a couple has a turning
effect, and the moment or torque of the couple is given
by r X F.
The following theorem is important.
F
Fig. 92
CHAP. 9]
PLANE MOTION OF RIGID BODIES
227
Theorem 9.5. Any system of forces which acts on a rigid body can be equivalently
replaced by a single force which acts at some specified point together with a suitable couple.
KINETIC ENERGY AND ANGULAR MOMENTUM ABOUT A FIXED AXIS
Suppose a rigid body is rotating about a fixed axis with
angular velocity <•> which has the direction of the axis AB
[see Fig. 93]. Then the kinetic energy of rotation is
given by
T = i/o> 2 (8)
where / is the moment of inertia of the rigid body about
the axis.
Similarly the angular momentum is given by
O = 7o> (9)
Fig. 93
MOTION OF A RIGID BODY ABOUT A FIXED AXIS
Two important methods for treating the motion of a rigid body about a fixed axis are
given by the following theorems.
Theorem 9.6: Principle of Angular Momentum. If A is the torque or the moment of
all external forces about the axis and O = I<a is the angular momentum, then
d
dt
(/•) = IS, = It
(10)
A =
where a is the angular acceleration.
Theorem 9.7: Principle of Conservation of Energy. If the forces acting on the rigid
body are conservative so that the rigid body has a potential energy V, then
T + V
i/ w 2 + V = E = constant
(«)
WORK AND POWER
Consider a rigid body % capable of rotating in a
plane about an axis O perpendicular to the plane, as
indicated in Fig. 94. If a is the magnitude of the torque
applied to the body under the influence of force F at
point A, the work done in rotating the body through
angle de is
dW = AdO (12)
and the instantaneous power developed is
dW
dt
cp 
= A.
(13)
Fig. 94
where w is the angular speed.
We have the following
Theorem 9.8. The total work done in rotating a rigid body from an angle 0i where
the angular speed is Wl to angle 2 where the angular speed is « a is the difference in the
kinetic energy of rotation at w x and o> 2 . In symbols,
J Ade = £/<of  £/o>f
(U)
228
PLANE MOTION OF RIGID BODIES
[CHAP. 9
IMPULSE. CONSERVATION OF ANGULAR MOMENTUM
The time integral of the torque
J»t 2
Adt
h
is called the angular impulse from time U to t 2 .
We have the following theorems.
(15)
Theorem 9.9. The angular impulse is equal to the change in angular momentum. In
symbols tf
I Adt
O2 — Oi
(16)
Theorem 9.10: Conservation of Angular Momentum. If the net torque applied to a
rigid body is zero, then the angular momentum is constant, i.e. is conserved.
THE COMPOUND PENDULUM
Let % [Fig. 95] be a rigid body which is free to oscillate
in a vertical plane about a fixed horizontal axis through O
under the influence of gravity. We call such a rigid body a
compound pendulum.
Let C be the center of mass and suppose that the angle
between OC and the vertical OA is 0. Then if h is the
moment of inertia of % about the horizontal axis through O,
M is the mass of the rigid body and a is the distance OC,
we have for the equation of motion,
H f— sm0
Jo
=
(17)
Fig. 95
For small oscillations the period of vibration is
P = 2ir\/Io/Mga (18)
The length of the equivalent simple pendulum is
I = hi Ma (19)
The following theorem is of interest.
Theorem 9.11. The period of vibration of a compound pendulum is a minimum when
the distance OC = a is equal to the radius of gyration of the body about the horizontal
axis through the center of mass.
GENERAL PLANE MOTION OF A RIGID BODY
The general plane motion of a rigid body can be considered as a translation parallel to
the plane plus a rotation about a suitable axis perpendicular to the plane. Two important
methods for treating general plane motion of a rigid body are given by the following
theorems.
Theorem 9.12: Principle of Linear Momentum. If r is the position vector of the center
of mass of a rigid body relative to an origin O, then
jfi (Mr) = Mr = F
(20)
where M is the total mass, assumed constant, and F is the net external force acting on
the body.
CHAP. 9]
PLANE MOTION OF RIGID BODIES
229
Theorem 9.13. Principle of Angular Momentum. If I c is the moment of inertia of the
rigid body about the center of mass, o> is the angular velocity and A c is the torque or
total moment of the external forces about the center of mass, then
= Tt^
>)
= 7 c"
(21)
Theorem 9.14. Principle of Conservation of Energy. If the external forces are conser
vative so that the potential energy of the rigid body is V, then
T + V = %mr 2 + i/ c o 2 + V = E = constant (22)
Note that imr 2 = \mv 2 is the kinetic energy of translation and i/ c *» 2 is the kinetic
energy of rotation of the rigid body about the center of mass.
INSTANTANEOUS CENTER.
SPACE AND BODY CENTRODES
Suppose a rigid body % moves parallel to a given
fixed plane, say the xy plane of Fig. 96. Consider
an x'y' plane parallel to the xy plane and rigidly
attached to the body.
As the body moves there will be at any time t
a point of the moving x'y' plane which is instan
taneously at rest relative to the fixed xy plane.
This point, which may or may not be in the body,
is called the instantaneous center. The line perpen
dicular to the plane and passing through the instan
taneous center is called the instantaneous axis.
O
Fig. 96
As the body moves, the instantaneous center also moves. The locus or path of the
instantaneous center relative to the fixed plane is called the space locus or space centrode.
The locus relative to the moving plane is called the body locus or body centrode. The motion
of the rigid body can be described as a rolling of the body centrode on the space centrode.
The instantaneous center can be thought of as that point about which there is rotation
without translation. In a pure translation of a rigid body the instantaneous center is
at infinity.
STATICS OF A RIGID BODY
The statics or equilibrium of a rigid body is the special case where there is no motion.
The following theorem is fundamental.
Theorem 9.15. A necessary and sufficient condition for a rigid body to be in equilibrium
is that
F = 0, A = (23)
where F is the net external force acting on the body and A is the net external torque.
PRINCIPLE OF VIRTUAL WORK AND D'ALEMBERT'S PRINCIPLE
Since a rigid body is but a special case of a system of particles, the principle of virtual
work and D'Alembert's principle [see page 171] apply to rigid bodies as well.
230
PLANE MOTION OF RIGID BODIES
[CHAP. 9
PRINCIPLE OF MINIMUM POTENTIAL ENERGY. STABILITY
At a position of equilibrium the net external force is zero, so that if the forces are
conservative and V is the potential energy,
F = vV = (24)
or in components,
?I = dV dV_
dx U ' dy U ' dz ~ °
(25)
In such case V is either a minimum or it is not a minimum. If it is a minimum the
equilibrium is said to be stable and a slight change of the configuration will restore the
body to its original position. If it is not a minimum the body is said to be in unstable
equilibrium and a slight change of the configuration will move the body away from its
original position. We have the following theorem.
Theorem 9.16. A necessary and sufficient condition for a rigid body to be in stable
equilibrium is that its potential energy be a minimum.
Solved Problems
RIGID BODIES
9.1. A rigid body in the form of a triangle ABC
[Fig. 97] is moved in a plane to position
DEF, i.e. the vertices A, B and C are car
ried to D, E and F respectively. Show that
the motion can be considered as a transla
tion plus a rotation about a suitable point.
Choose a point G on triangle ABC which cor
responds to the point H on triangle DEF. Perform
the translation in the direction GH so that triangle
ABC is carried to A'B'C. Using H as center of
rotation perform the rotation of triangle A'B'C
through the angle 9 as indicated so that A'B'C is
carried to DEF. Thus the motion has been accom
plished by a translation plus a rotation.
Fig. 97
9.2.
Give an example to show that finite rotations cannot be represented by vectors.
Let A x represent a rotation of a body [such as the rectangular parallelepiped of Fig. 98(a)]
about the x axis while A y represents a rotation about the y axis. We assume that such rotations
take place in a positive or counterclockwise sense according to the right hand rule.
R
V
tf
r (6)
Fig. 98
CHAP. 9]
PLANE MOTION OF RIGID BODIES
231
In Fig 98(a) we start with the parallelepiped in the indicated position and perform the rotation
A x about the x axis as indicated in Fig. 98(6) and then the rotation about the » axis m indi
cated in Fig. 98(c). Thus Fig. 98(c) is the result of the rotation A x + A y on Fig. 98(o).
In Fig 99(a) we start with the parallelepiped in the same position as in Fig. 98(a), but this
time, we first perform the rotation A y about the y axis as indicated in Fig. 99(6) and then the
rotation A x about the x axis as indicated in Fig. 99(c). Thus Fig. 99(c) is the result of the
rotation A y + A x on Fig. 99(a).
Since the position of the parallelepiped of Fig. 98(c) is not the same as that of Fig. 99(c),
we conclude that the operation A x + A y is not the same as A y + A x . Thus the commutative
law is not satisfied, so that A x and A y cannot possibly be represented by vectors.
MOMENTS OF INERTIA
9.3. Two particles of masses m x and ra 2 respectively are connected by a rigid massless
rod of length a and move freely in a plane. Show that the moment of inertia of the
system about an axis perpendicular to the plane and passing through the center of
mass is iacl 2 where the reduced mass
ix — m\m<2.l{m\ + W2).
Let r x be the distance of mass mj from the
center of mass C. Then a — r x is the distance
of mass m 2 from C. Since C is the center
of mass,
m l r 1 = m 2 (a — r t ) from which r
m^a
mj + W2
Fig. 910
and a — r t
w^a
!»! + W^
Thus the moment of inertia about an axis through C is
m x r\ + m 2 (a — r x ) 2 = m.
m 2 a
. + m 2 ( : /
m x + m 2 / \ m i + m 2/
m x a \ 2 _ wii%
fllj+W^
a 2 = ndP
9.4. Find the moment of inertia of a solid circular cylinder of radius a, height h and
mass M about the axis of the cylinder.
Method 1, using single integration.
Subdivide the cylinder, a cross section of which appears
in Fig. 911, into concentric rings one of which is the ele
ment shown shaded. The volume of this element is
(Area) (thickness) = (2irrdr)(h) = 2rrrh dr
and the element of mass is dm = 2varh dr.
The moment of inertia of dm is
r 2 dm — 2iror a h dr
where a is the density, and thus the total moment of inertia is
J id
2war s h dr = ^waha A
(1)
Fig. 911
232
PLANE MOTION OF RIGID BODIES
[CHAP. 9
Then since the mass is
we find / = Ma 2 .
M
2worh dr
aira 2 h
Method 2, using double integration.
Using polar coordinates (r, 8), we see from Fig. 912 that
the moment of inertia of the element of mass dm distant
r from the axis is
r 2 dm = r 2 ahrdrdo = ahr* dr do
since hr dr do is the volume element and a is the mass per
unit volume (density). Then the total moment of inertia is
J»2ir s»a
I ahr^drdo = \irohcfi
(1)
0=0 "r=0
The mass of the cylinder is given by
M
s»2tt pa
^6=0 ^r=0
ahr dr do = aira 2 h
Fig. 912
which can also be found directly by noting that the volume of the cylinder is ira 2 h. Dividing
equation (1) by (2), we find I/M = %a 2 or I = \Ma 2 .
9.5. Find the radius of gyration, K, of the cylinder of Problem 9.4.
Since K 2 = I/M = \a 2 , K = a/y/2 = \a^2.
96. Find the (a) moment of inertia and (b) radius of gyration of a rectangular plate
with sides a and b about a side.
Method 1, using single integration.
(a) The element of mass shaded in Fig. 913 is ab dx, and its moment of inertia about the y axis is
(ab dx)x 2 = abx 2 dx. Thus the total moment of inertia is
J ia
abx 2 dx = %aba*
Since the total mass of the plate is M = aba, we have I/M = Ja 2 or / = $Ma 2 .
.(b) K 2 = I/M = %a 2 or K = a/y[S = %ayfz.
y
dx — •»
1
b
1
Fig. 913
6 *
dm = a dy dx
Fig. 914
Method 2, using double integration.
Assume the plate has unit thickness. If dm = adydx is an element of mass [see Fig. 914],
the moment of inertia of dm about the side which is chosen to be on the y axis is x 2 dm = ax 2 dy dx.
Then the total moment of inertia is
Xa s\h
I ax 2 dydx = j^aba?
The total mass of the plate is M = aba. Then, as in Method 1, we find / = \Ma 2 and K = $ayfs.
CHAP. 91
PLANE MOTION OF RIGID BODIES
231
9.7. Find the moment of inertia of a right circular cone of height h and radius a
about its axis.
Method 1, using single integration.
The moment of inertia of the circular cylindrical
disc one quarter of which is represented by PQR in
Fig. 915 is, by Problem 9.4,
\{irr 2 o dz)(r 2 ) = ±iror*dz
since this disc has volume vr 2 dz and radius r.
From Fig. 915,
h — z
r
— or r
a
h a ™\ h
Then the total moment of inertia about the z axis is
h
dz = JL
Also,
M =
dz —
^jjTra 4 <rh
±ira 2 ha
Fig. 915
Thus / = ^Ma 2 .
Method 2, using triple integration.
Subdivide the cone, one quarter of which is
shown in Fig. 916, into elements of mass dm as indi
cated in the figure.
In cylindrical coordinates (r, 8, z) the element of
mass dm of the cylinder is dm — or dr de dz where
a is the density.
The moment of inertia of dm about the z axis is
r 2 dm = or 3 dr de dz
As in Method 1,
h — z
h
ha \ a
Then the total moment of inertia about the z axis is
>h(a — r)/a
Fig. 916
The total mass of the cone is
J»27r pa s»n,(a — r)/a,
0=0 J r =0 ^z=0
is
*S0=O ^r=0 *^2=0
dr de dz — ^iraAoh
■Jiia—rj/r
M =
or dr de dz
^■naPho
which can be obtained directly by noting that the volume of the cone is \tra 2 h.
Thus /
^Ma 2 .
9.8. Find the radius of gyration K of the cone of Problem 9.7.
K 2 = I/M = ^a 2 and K = ay/^ = ^ay/SO.
THEOREMS ON MOMENTS OF INERTIA
9.9. Prove the parallel axis theorem [Theorem 9.3, page 226].
Let OQ be any axis and ACP a parallel axis through the centroid C and distant 6 from OQ.
In Fig. 917 below, OQ has been chosen as the z axis so that AP is perpendicular to the xy plane
at P.
234
PLANE MOTION OF RIGID BODIES
[CHAP. 9
If b x is a unit vector in the direction OP,
then the vector OP is given by
b = 6bi (1)
where b is constant and is the distance between
axes.
Let r„ and x' v be the position vectors of
mass m„ relative to O and C respectively. If r
is the position vector of C relative to O then
we have
r v = x' v + f (jg)
The total moment of inertia of all masses
m v about axis OQ is
N
/ = 2 m^.bi)* (3) Fig. 917
The total moment of inertia of all masses m v about axis ACP is
N
Ic = 2 m^b^
Then using (2) we find
N N
I = 2 m^.bi) 2 = 2 mX'bi + fbi) 2
K=l V=l
N N N
= 2 m^b^ + 22 m^.b^Cfbj) + 2 ^(fb^
V=l V = l 1>=1
/ N \ N
= / c + 26 ( 2 m, < ) . b x + 6 2 2 m v = I c + Mb*
N N
since i'b l = b, ^ m v = M and 2 whrj = [Problem 7.16, page 178].
r=l v=l
The result is easily extended to continuous mass systems by using integration in place of
summation.
(•4)
9.10. Use the parallel axis theorem to find the moment of inertia
of a solid circular cylinder about a line on the surface of
the cylinder and parallel to the axis of the cylinder.
Suppose the cross section of the cylinder is represented as in
Fig. 918. Then the axis is represented by C, while the line on the
surface of the cylinder is represented by A.
If a is the radius of the cylinder, then by Problem 9.4 and the
parallel axis theorem we have
j A = I c + Ma? = ±Ma 2 + Ma*  f Ma*
Fig. 918
9.11. Prove the perpendicular axes theorem [Theorem
9.4, page 226].
Let the position vector of the particle with mass m v
in the xy plane be
r„ = x v i + yj
[see Fig. 919]. The moment of inertia of m„ about the
z axis is w^r„ 2 .
Then the total moment of inertia of all particles *
about the z axis is
Fig. 919
CHAP. 9]
PLANE MOTION OF RIGID BODIES
235
N N
h =. 2«hrr„« = U(^ 2 + a
v = \ v = l
= 2 m v xl + 2 m,!^ = I~ + I„
v=l v =i
where I x and J B are the total moments of inertia about the x axis and y axis respectively.
The result is easily extended to continuous systems.
9.12. Find the moment of inertia of a rectangular plate with sides a and b about an axis
perpendicular to the plate and passing through a vertex.
Choose the rectangular plate [see Fig. 920] in the
xy plane with sides on the x and y axes. Choose the
z axis perpendicular to the plate at a vertex.
From Problem 9.6 we have for the moments of
inertia about the x and y axes,
I x = $Mb*, I y = ^Ma2
Then by the perpendicular axes theorem the moment
of inertia about the z axis is
I z = I x + I y = £M(&2 + a 2)
= $M(a 2 + b 2 )
Fig. 920
COUPLES
9.13. Prove that a force acting at a point of a rigid body can be equivalents replaced
by a single force acting at some specified point together with a suitable couple.
Let the force be F t acting at point P t as in
Fig. 921. If Q is any specified point, it is seen that
the effect of F 4 alone is the same if we apply two
forces t t and — f x at Q.
In particular if we choose f t = —F u i.e. if f
has the same magnitude as Fj but is opposite in
direction, we see that the effect of F x alone is the
same as the effect of the couple formed by F t and
f i  ~ F i [which has moment r,XF,] together with the
force — fj = F 1#
Fig. 921
9.14. Prove Theorem 9.5 page 227: Any system of forces which acts on a rigid body can
witlTsS cXr ^ 3 Single f0Ke WhiCH ^ at — «"*" "* « a er
By Problem 9.13 we can replace the force F„
at P„ by the force P„ at Q plus a couple of moment
r„ X F„. Then the system of forces P lf F 2 , . . . , F N at
points P lt P 2 ,...,P N can be combined into forces
Fj, F 2 , . . . , F N at Q having resultant
F = F 1 + F 2 + • • • + Fn
together with couples having moments
'iXFj, r 2 XF 2 , ..., r N XF N
which may be added to yield a single couple. Thus
the system of forces can be equivalent^ replaced by
the single force F acting at Q together with a couple
.Py
Fig. 922
236
PLANE MOTION OF RIGID BODIES
[CHAP. 9
KINETIC ENERGY AND ANGULAR MOMENTUM
9.15. If a rigid body rotates about a fixed axis with
angular velocity ©, prove that the kinetic energy
of rotation is T = ^/w 2 where / is the moment of
inertia about the axis.
Choose the axis as AB in Fig. 923. A particle P of
mass m v will rotate about the axis with angular speed «.
Then it will describe a circle PQRSP with linear speed
v v — ur v where r v is its distance from axis AB. Thus its
kinetic energy of rotation about AB is ^m v v% = ±m v w 2 r 2 ,
and the total kinetic energy of all particles is
T = 2 ^rn v ^rl
I ( 2 w„r 2 )<o 2
v = l
= ¥" 2
where / = 2 w„ r 2 is the moment of inertia about AB.
v = l
The result could also be proved by using integration in
place of summation.
Fig. 923
9.16. Prove that the angular momentum of the rigid body of Problem 9.15 is O = /».
The angular momentum of particle P about axis AB is m v r*m. Then the total angular
momentum of all particles about axis AB is
N / N \
a  2 m v r 2 , o> = ( 2 «iv r 2 ) » = /<*
where I — 2 w„rf is the moment of inertia about AB.
v = l
The result could also be proved by using integration in place of summation.
MOTION OF A RIGID BODY ABOUT A FIXED AXIS
9.17. Prove the principle of angular momentum for a rigid body rotating about a fixed
axis [Theorem 9.6, page 227].
By Problem 7.12, page 176, since a rigid body is a special case of a system of particles,
A = dtl/dt where A is the torque or moment of all external forces about the axis and Q is the
total angular momentum about the axis.
Since O = /<■> by Problem 9.16,
a _ d ir \ — T da
A  di (Ia) ' ! Tt
/i.
9.18. Prove the principle of conservation of energy for a rigid body rotating about a
fixed axis [Theorem 9.7, page 227] provided the forces acting are conservative.
The principle of conservation of energy applies to any system of particles in which the forces
acting are conservative. Hence in particular it applies to the special case of a rigid body rotating
about a fixed axis. If T and V are the total kinetic energy and the potential energy, we thus have
T + V = constant
E
Using the result of Problem 9.15, this can be written ^Iu> 2 + V  E.
CHAP. 9]
PLANE MOTION OF RIGID BODIES
237
WORK, POWER AND IMPULSE
9.19. Prove equation (12), page 227, for the work done in rotating a rigid body about a
fixed axis.
Refer to Fig. 94, page 227. Let the angular velocity of the body be « = „k where k is a
unit vector m the direction of the axis of rotation. The work done by F is
fJr
dW = Ftfr  Ff t dt = Fvd* = F.(.Xr)<ft
= (rXF)»(lf = A • m dt = A w dt = Ade
where in the last two steps we use A = Ak, o> = «k and a  de/dt.
9.20. Prove equation (18), page 227, for the power developed.
From Problem 9.19 and the fact that de/dt = a ,
¥ = dW/dt = A de/dt = Aw
9.21. Prove Theorem 9.8, page 227.
deJl e dt h Zlhte Id ° /dt S ° tHat A = Idu/dt  Then from Problem 9  19 ^d the fact that
ol Ade = \ l %» dt = l t '*• = V4M
9.22. Prove Theorem 9.9, page 228: The angular impulse is equal to the change in
angular momentum.
f\dt = C
J tl J h dt
dt = Q 2  O x
9.23. Prove Theorem 9.10, page 229, on the conservation of angular momentum if the net
torque is zero.
From Problem 9.22, if A = then Q 2 = Q v
THE COMPOUND PENDULUM
9.24. Obtain the equation of motion (17), page 228, for
a compound pendulum.
Method 1.
Suppose that the vertical plane of vibration of the
pendulum is chosen as the xy plane [Fig. 924] where
the z axis through origin O is the horizontal axis of
suspension.
Let point C have the position vector a relative to O.
Since the body is rigid, a = a is constant and is the
distance from O to C.
The only external force acting on the body is its
weight Mg = Mgj acting vertically downward. Thus
we have
A = total external torque about z axis
= a X Mg = a X Mgj = aMg sin e k (1)
where k is a unit vector in the positive z direction [out of the plane of the paper toward the reader].
Also, the instantaneous angular velocity is
_ i de . .
«  <ok = ~^k = *k
Fig. 924
so that if I is the moment of inertia about the z axis
238 PLANE MOTION OF RIGID BODIES [CHAP. 9
O = angular momentum about z axis = /<><* — — M k
Substituting from (1) and (2) into A = dil/dt,
aMg sin fk = r(7o*k) or IS + ^^ sin * = (5)
Method 2.
The force Mg = — Mgj is conservative, so that the potential energy V is such that
_ vv = «E, _ *E j _ £ k = _»„ or £ = 0, f = *,, £ =
da; fly dz 9» 92/ Sz
from which V = Mgy + c = —Mga cos o + c (4)
since 1/ = —a cos d. This could be seen directly since y = —a cos e is the height of C below the
x axis taken as the reference level.
By Problem 9.15, the kinetic energy of rotation is ±I u 2  ^I <> 2  Then the principle of
conservation of energy gives
T + V  / o 2  Mga cos = constant = E (5)
Differentiating equation (5) with respect to t,
1 6 'S + Mga sin e —
or, since is not identically zero, I e + Mga sin $ — as required.
9.25. Show that for small vibrations the pendulum of Problem 9.24 has period
P = 27T\/Mga/h.
For small vibrations we can make the approximation sin = so that the equation of motion
becomes „,
%+*¥*, = o a)
Then, as in Problem 4.23, page 102, we find that the period is P = 2jry/I /Mga.
9.26. Show that the length I of a simple pendulum equivalent to the compound pendulum
of Problem 9.24 is I = I of Ma.
The equation of motion corresponding to a simple pendulum of length I suspended vertically
from O is [see Problem 4.23, equation (2), page 102]
6 + j sin e  (0
Comparing this equation with (1) of Problem 9.25, we see that I = I /Ma.
GENERAL PLANE MOTION OF A RIGID BODY
9.27. Prove the principle of linear momentum, Theorem 9.12, page 228, for the general
plane motion of a rigid body.
This follows at once from the corresponding theorem for systems of particles [Theorem 71,
page 167], since rigid bodies are special cases.
9.28. Prove the principle of angular momentum, Theorem 9.13, page 229, for general
plane motion of a rigid body.
This follows at once from the corresponding theorem for systems of particles [Theorem 74,
page 168], since rigid bodies are special cases.
CHAP. 9]
PLANE MOTION OF RIGID BODIES
239
9.29. A solid cylinder of radius a and
mass M rolls without slipping down
an inclined plane of angle a. Show
that the acceleration is constant and
equal to f g sin a.
Suppose that initially the cylinder has
point O in contact with the plane and that
after time t the cylinder has rotated
through angle [see Fig. 925].
The forces acting on the cylinder at
time t are: (i) the weight Mg acting verti
cally downward at the center of mass C;
(ii) the reaction R of the inclined plane act
ing perpendicular to the plane; (iii) the
frictional force f acting upward along the
incline.
Fig. 925
Choose the plane in which motion takes place as the xy plane, where the * axis is taken as
positive down the incline and the origin is at O.
If r is the position of the center of mass at time t, then by the principle of linear momentum,
Mr = Mg + R + f
But g = g sin a i  g cos a j, R = Rj, f = /i. Hence (1) can be written
Mr* = (Mg sin a  f)\ + (R  Mg cos «)j
The total external torque about the horizontal axis through the center of mass is
A = OXMg + OXR + CBXf = CBXf ^ (aj)X(/i) =  a /k
The total angular momentum about the horizontal axis through the center of mass is
O = / c o, = 7 c (*k) = 7 c *k
where I c is the moment of inertia of the cylinder about this axis.
Substituting (3) and U) into A = dCl/dt, we find a/k = 7 c Vk or I c e = af.
Using r = xi + yj in (2), we obtain
M x = Mg sin a ~ f, My = R  Mg cos a
W
(2)
(3)
U)
(5)
Nowjf there is no slipping, x = ae or = x/a. Similarly, since the cylinder remains on the
incline, y = 0; hence from (5), R = Mg cos a.
Using e = x/a in I c e = af, we have f = I c '£/a*. From Problem 9.4, I c = JLMa* Then
substituting / = iMx into the first equation of (5), we obtain x = \g sin a as required.
9.30. Prove that in Problem 9.29 the coefficient of friction must be at least * tan«.
The coefficient of friction is n — f/R.
From Problem 9.29 we have / = \M x = ±Mg sin a and B = Mg cos a. Thus in order that
slipping will not occur, n must be at least f/R = 1 tan a.
9.31. (a) Work Problem 9.29 if the coefficient of friction between the cylinder and inclined
plane is ^ and (b) discuss the motion for different values of /*.
(a) In equation (5) of Problem 9.29, substitute / = pR = ^Mg cos a and obtain
*** = £f(sin a — ix cos a)
Note that in this case the center of mass of the cylinder moves in the same manner as
a particle sliding down an inclined plane. However, the cylinder may slip as well as roll.
240
PLANE MOTION OF RIGID BODIES
[CHAP. 9
The acceleration due to rolling is
a 9 =
a 2 f a 2 /iMg cos a
= 2ng cos a.
Ic \M&
The acceleration due to slipping is x — ae = g(sin a — 3/* cos a).
(b) If (sin a — 3/i cos a) > 0, i.e. u < ^ tan a, then slipping will occur. If (sin a — 3fi cos a) ^ 0,
i.e. /* = ^ tan a, then rolling but no slipping will occur. These results are consistent with
those of Problem 9.30.
9.32. Prove the principle of conservation of energy [Theorem 9.14, page 229].
This follows from the corresponding theorem for systems of particles, Theorem 77, page 169.
The total kinetic energy T is the sum of the kinetic energy of translation of the center of mass
plus the kinetic energy of rotation about the center of mass, i.e.,
T = imr 2 + / c <o 2
If V is the potential energy, then the principle of conservation of energy states that if E
is a constant,
T + V = mf 2 + £/ c <o 2 + V = E
9.33. Work Problem 9.29 by using the principle of conservation of energy.
The potential energy is composed of the potential energy due to the external forces [in this
case gravity] and the potential energy due to internal forces [which is a constant and can be
omitted]. Taking the reference level as the base of the plane and assuming that the height of
the center of mass above this plane initially and at any time t to be H and h respectively, we have
IMf 2 + l/ c <o 2 + Mgh  MgH
or, using H — h = x sin a
and r 2 = x 2 + y 2 = x 2 since y = 0,
\Mx 2 + ^7 c w 2 = Mgx sin a
Substituting w = — x/a and I c — \Ma 2 , we find x 2 = \gx sin a. Differentiating with respect to
t, we obtain
2x x = §gx sin a or x = f g sin a
INSTANTANEOUS CENTER. SPACE AND BODY CENTRODES
9.34. Find the position vector of the instantaneous center for a rigid body moving
parallel to a given fixed plane.
Choose the XY plane of Fig. 926 as the fixed
plane and the xy plane as the plane attached to
and moving with the rigid body %. Let point P of
the xy plane [which may or may not be in the
rigid body] have position vectors R and r relative
to the XY and xy planes respectively. If v and
v^ are the respective velocities of P and A relative
to the XY system,
v = v A + »Xr = V A + o» X (R — R A ) (1)
where R A is the position vector of A relative to O.
If P is to be the instantaneous center, then v =
so that
• X(RR A ) = v A («)
Multiplying both sides of (2) by o> X and using (7), Fig. 926
page 5,
«{«»(R — R A )} — (R — R A )(«»«) = «Xv A
Then since » is perpendicular to R — R A , this becomes
(R  R A )w 2 = » X v A or R = R A +
« X v A
(3)
CHAP. 9]
PLANE MOTION OF RIGID BODIES
241
9.35. A cylinder moves along a horizontal plane. Find the (a) space centrode, (o) body
centrode. Discuss the case where slipping may occur.
(a)
The general motion is one where both rolling
and slipping may occur. Suppose the cylinder
is moving to the right with velocity v A [the
velocity of its center of mass] and is rotating
about A with angular velocity «.
Since o> = — wk and v A = v A i, we have
• x v a = wv A j so that (3) of Problem 9.34
becomes . . .
_ _ (« v a)j „ v A#
R = R A  — X~ = R A 3
In component form,
Fig. 927
Xi + Yj  X A i + oj  (v A /«)j x>r X = X A , Y = a  v A /a
Thus the instantaneous center is located vertically above the point of contact of the cylinder
with the ground and at height a — v A /u above it.
Then the space centrode is a line parallel to the horizontal and at distance a — v A /u
above it. If there is no slipping, then v A = aa and the space centrode is the X axis while
the instantaneous center is the point of contact of the cylinder with the X axis.
(b) The body centrode is given by r = v /o, or a circle of radius v /u. In case of no slipping,
v = ao3 and the body centrode is the circumference of the cylinder.
9.36. Solve Problem 9.29 by using the instantaneous center.
By Problem 9.35, if there is no slipping then
the point of contact P of the cylinder with the
plane is the instantaneous center. The motion of
P is parallel to the motion of the center of mass,
so that we can use the result of Problem 7.86(c),
page 191.
The moment of inertia of the cylinder about
P is, by the parallel axis theorem, ^Ma 2 + Ma 2 =
2 Ma 2 . The torque about the horizontal axis through
P is Mga sin a. Thus
^{lMa 2 'e) = Mga sine
or e = r^smfl
6a
Since x = ae, the acceleration is x = f g sin 9.
Fig. 928
STATICS OF A RIGID BODY
9.37. A ladder of length I and weight Wi has one end
against a vertical wall which is frictionless and
the other end on the ground assumed horizontal.
The ladder makes an angle a with the ground.
Prove that a man of weight W m will be able to
climb the ladder without having it slip if the
coefficient of friction /* between the ladder and
the ground is at least
Wm + Wl
Wm+Wr
COt a.
Let the ladder be represented by AB in Fig. 929 and
choose an xy coordinate system as indicated.
Fig. 929
242
PLANE MOTION OF RIGID BODIES
[CHAP. 9
The most dangerous situation in which the ladder would slip occurs when the man is at the
top of the ladder. Hence we would require that the ladder be in equilibrium in such case.
The forces acting on the ladder are: (i) the reaction R x = R t i of the wall; (ii) the weight
W m = — W m j of the man; (iii) the weight W £ = — W t j of the ladder concentrated at C, the center of
gravity; (iv) the reaction R 2 = R 2 j of tn © ground; (v) the friction force f = — /i.
For equilibrium we require that
F = 0, A = (I)
where F is the total external force on the ladder and A the total external torque taken about a
suitable axis which we shall take as the horizontal axis through A perpendicular to the xy plane.
We have
F = R x + W m + W, + R 2 + f = (R 1 f)i + (W m W t + R 2 )j =
if JBj — / = and W m  W, + R 2 = (2)
Also, A = (0) X Bi + (0) X W m + (AC) XW,+ (AB) X B 2 + (AB) X f
= (0) X (flji) + (0) X ( W m j) + ($1 cos a i  ±1 sin a j) x (TFJ)
+ (I cos a i — I sin a j) X (R 2 j) + (I cos a i — I sin a j) X (— /i)
= —tyWi cos a k + IR 2 cos a. k — If sin a k =
—^Wi COS a + R 2 COS a — f sin a =
if
Solving simultaneously equations (2) and (3), we find
/ = R l = (W TO + W l )cot«
and
R, = W m + W,
Then the minimum coefficient of friction necessary to prevent slipping of the ladder is
cot a
/ _ w m + jW t
Ra
w m + w l
(3)
MISCELLANEOUS PROBLEMS
9.38. Two masses mi and m 2 are connected by an inextensible string of negligible mass
which passes over a frictionless pulley of mass M, radius a and radius of gyration K
which can rotate about a horizontal axis through C perpendicular to the pulley.
Discuss the motion.
Choose unit vectors i and j in the plane of rotation
as shown in Fig. 930.
If we represent the acceleration of mass m x by Aj,
then the acceleration of mass ra 2 is — Aj.
Choose the tensions T x and T 2 in the string as shown
in the figure. By Newton's second law,
CO
(2)
(3)
U)
mg
wiAj = T x + m x g = — TJ + m^j
mgAj = T 2 + m 2 g = T 2 j + m^j
Thus m x A = mxQ — T lt m 2 A — T 2 — m 2 g
or T t = m^gA), T 2 = m 2 {g + A)
The net external torque about the axis through C is
A = (oi) X (TJ) + (ai) X (r^) = a(7\  T 2 )k
The total angular momentum about O is
CI = I c » = 7 c cok = IcOk
Since A = dil/dt, we find from (5) and (6),
a{T x T 2 ) = I c 'o = MK*$
Fig. 930
(5)
(6)
(7)
CHAP. 9]
PLANE MOTION OF RIGID BODIES
243
If there is no slipping about the pulley, we also have
A — a'e
Using (8) in (7),
Using (4) in (9),
Tx ~ T 2
A =
MK 2
(wn  m 2 )g
m 1 + m 2 + MkVa?
(8)
(9)
(10)
Thus the masses move with constant acceleration given in magnitude by (10). Note that if M — 0,
the result (10) reduces to that of Problem 3.22, page 76.
9.39. Find the moment of inertia of a solid sphere about a diameter.
Let O be the center of the sphere and AOB be the
diameter about which the moment of inertia is taken
[Fig. 931]. Divide the sphere into discs such as QRSTQ
perpendicular to AOB and having center on AOB at P.
Take the radius of the sphere equal to a, OP = z,
SP = r and the thickness of the disc equal to dz. Then
by Problem 9.4 the moment of inertia of the disc about
AOB is
\(trr 2 o dz)r 2  ^var^dz (1)
From triangle OSP, r 2 = a 2  z 2 .
the total moment of inertia is
Substituting into (1),
= f
\ira(a 2  z 2 ) 2 dz
The mass of the sphere is
M
f_
•gwa 3 a
7ra(a 2  z 2 ) dz =
2= —a
which could also be seen by noting that the volume of the sphere is ova 3 .
($)
From (2) and (S) we have I/M = \a 2 ox I  %Ma 2 .
9.40. A cube of edge s and mass M is suspended vertically from one of its edges, (a) Show
that the period for small vibrations is P = 2v ^2y/s/Sg. (b) What is the length of
the equivalent simple pendulum?
(a)
Since the diagonal of a square of side 8 has length V» 2 + s 2 = 8^/2,
the distance OC from axis O to the center of mass is ^8^/2.
The moment of inertia / of a cube about an edge is the same as
that of a square plate about a side. Thus by Problem 9.6,
/ = %M(s 2 + s 2 ) = §Ms 2 .
Then the period for small vibrations is, by Problem 9.25,
P = 2v *sl%Ms 2 /[Mg($ 8 V2)] = 2wyf2y[slZg~
(b) The length of the equivalent simple pendulum is, by Problem 9.26,
I = %MsV[M(%sV2)] = fv^s
Fig. 932
9.41. Prove Theorem 9.11, page 228: The period of small vibrations of a compound
pendulum is a minimum when the distance OC = a is equal to the radius of gyration
of the body about a horizontal axis through the center of mass.
If I c is the moment of inertia about the center of mass axis and I is the moment of inertia
about the axis of suspension, then by the parallel axis theorem we have
244
PLANE MOTION OF RIGID BODIES
[CHAP. 9
7 = i c + Ma 2
Then the square of the period for small vibrations is given by
p2 =
4fl 2 /p _ 4ir 2 / h
+ a
2/Kr
+ a
Mga g \Ma J g
where K* = I c / M is the square of the radius of gyration about the center of mass axis.
Setting the derivative of P 2 with respect to a equal to zero, we find
da^ ' g \ a?
from which a  K c . This can be shown to give the minimum value since d 2 (P 2 )/da 2 < 0. Thus
the theorem is proved.
The theorem is also true even if the vibrations are not assumed small. See Problem 9.147.
=
9.42. A sphere of radius a and mass m rests on top of a fixed rough sphere of radius b.
The first sphere is slightly displaced so that it rolls without slipping down the second
sphere. Where will the first sphere leave the second sphere?
Let the xy plane be chosen so as to pass
through the centers of the two spheres, with the
center of the fixed sphere as origin O [see Fig.
933]. Let the position of the center of mass C of
the first sphere be measured by angle 0, and sup
pose that the position vector of this center of
mass C with respect to O is r. Let r l and 9 X be
unit vectors as indicated in Fig. 933.
Resolving the weight W = — mgj into com
ponents in directions r t and 9 lt we have [compare
Problem 1.43, page 24]
W = (WT^ + (W •*!)#!
= (mgjrj)^ + (mgj • 9 t ) 9 X
= —mg sin o r t — mg cos e 9 t
The reaction force N and frictional force f
are N = Nr lt f = f9 v Using Theorem 9.12,
page 228, together with the result of Problem 1.49,
page 26, we have
F = ma = m[(r — r'o 2 )r x + (r'e + 2ro)9 1 ]
= W + N + £
= (N — mg sin 0)^ + (/ — mg cos
re*)  N — mg sine, m(r'o + 2re) = f
Fig. 933
from which m(r
Since r = a + b [the distance of C from O], these equations become
= /
mg cos e
mg cos 6
(1)
—m(a + b)'e 2 = N — mg sin e, m(a + b)
We now apply Theorem 9.13, page 229. The total external torque of all forces about the
center of mass C is [since W and N pass through C],
A = (ar,) X f = (ar 1 )X(f9 1 ) = a/k
Also, the angular acceleration of the first sphere about C is
d 2
a = ^(0 + v^)k = (<£ + vOk
Since there is only rolling and no slipping it follows that arc AP equals arc BP, or b$ = af.
Then </> = v/2  e and ^ = (b/a)(ir/2 — o), so that
= (0 + vOk
{ 6 ~ $
a
^±* )*k
CHAP. 9] PLANE MOTION OF RIGID BODIES 245
Since the moment of inertia of the first sphere about the horizontal axis of rotation through C
is I — &ma 2 , we have by Theorem 9.13,
A = la, afk = wa2(^i^ )Vk or / = im(a+b)e
Using this value of / in the second equation of (1), we find
5g
6 = "7(^+6) C0S6 W
Multiplying both sides by $ and integrating, we find after using the fact that e = at t = or
e = r/2,
Using (*) in the first of equations (J), we find A/" = lwflr(17 sin  10). Then the first sphere
leaves the second sphere where N = 0, i.e. where 6 — sin* 10/17.
Supplementary Problems
RIGID BODIES
9.43. Show that the motion of region % of Fig. 934 can
be carried into region %' by means of a translation
plus a rotation about a suitable point.
9.44. Work Problem 9.1, page 230, by first applying a
translation of the point A of triangle ABC.
9.45. If A X ,A V ,A Z represent rotations of a rigid body about
the x, y and z axes respectively, is it true that the
associative law applies, i.e. is A x + (A y + A z ) =
(A x + A y ) + A z t Justify your answer. Fig. 934
MOMENTS OF INERTIA
9.46. Three particles of masses 3, 5 and 2 are located at the points (1, 0, 1), (2, 1, 3) and (2, 2, 1)
respectively. Find (a) the moment of inertia and (6) the radius of gyration about the x axis.
Ans. 71
9.47. Find the moment of inertia of the system of particles in Problem 9.46 about (a) the y axis,
(fe) the z axis. Ans. (a) 81, (6) 44
9.48. Find the moment of inertia of a uniform rod of length I about an axis perpendicular to it and
passing through (a) the center of mass, (b) an end, (c) a point at distance 1/4 from an end.
Ans. (a) r&MP, (b) j^MP, (c) ^MP
9.49. Find the (a) moment of inertia and (b) radius of gyration of a square of side a about a diagonal.
Ans. (a) ^Ma*, (b) ^ay/S
9.50. Find the moment of inertia of a cube of edge a about an edge. Ans. § Ma 2
9.51. Find the moment of inertia of a rectangular plate of sides a and 6 about a diagonal.
Ans. ±Ma 2 b 2 /(a 2 + b 2 )
9.52.
Find the moment of inertia of a uniform parallelogram of sides a and 6 and included angle a
about an axis perpendicular to it and passing through its center. Ans. J*M(a 2 + b 2 ) sin 2 a
246
PLANE MOTION OF RIGID BODIES
[CHAP. 9
9.53.
9.54.
9.55.
9.56.
9.57.
9.58.
9.59.
Find the moment of inertia of a cube of side a about a diagonal.
Find the moment of inertia of a cylinder of radius a and height h
about an axis parallel to the axis of the cylinder and distant b from
its center. Ans. ^M(a 2 + 2b 2 )
A solid of constant density is formed from a cylinder of radius a
and height h and a hemisphere of radius a as shown in Fig. 935.
Find its moment of inertia about a vertical axis through their
centers. Ans. M(2a 3 + 15a 2 h)/{10a + 15h)
Work Problem 9.55 if the cylinder is replaced by a cone of radius a
and height h.
Find the moment of inertia of the uniform solid region bounded
by the paraboloid cz = x 2 + y 2 and the plane z = h about the
z axis. Ans. ^Mch
Fig. 935
How might you define the moment of inertia of a solid about (a) a point, (6) a plane? Is there
any physical significance to these results? Explain.
Use your definitions in Problem 9.58 to find the moment of inertia of a cube of side a about
(a) a vertex and (6) a face. Ans. (a) Ma 2 , (b) ilfa 2
KINETIC ENERGY AND ANGULAR MOMENTUM
9.60. A uniform rod of length 2 ft and mass 6 lb rotates with angular speed 10 radians per second
about an axis perpendicular to it and passing through its center. Find the kinetic energy of
rotation. Ans. 100 lb ft 2 /sec 2
9.61. Work Problem 9.60 if the axis of rotation is perpendicular to the rod and passes through an end.
Ans. 400 lb ft 2 /sec 2
9.62.
9.63.
9.64.
9.65.
9.66.
9.67.
A hollow cylindrical disk of radius a and mass M rolls along a horizontal plane with speed v.
Find the total kinetic energy. Ans. Mv 2
Work Problem 9.62 for a solid cylindrical disk of radius a.
Ans. &Mv 2
A flywheel having radius of gyration 2 meters and mass 10 kilograms rotates at angular
speed of 5 radians/sec about an axis perpendicular to it through its center. Find the kinetic
energy of rotation. Ans. 1000 joules
Find the angular momentum of (a) the rod of Problem 9.60 (6) the flywheel of Problem 9.64.
Ans. (a) 5 lb ft 2 /sec, (b) 200 kg m 2 /sec
Prove the result of (a) Problem 9.15, page 236, (6) Problem 9.16, page 236, by using integration
in place of summation.
Derive a "parallel axis theorem" for (a) kinetic energy and (6) angular momentum and explain
the physical significance.
MOTION OF A RIGID BODY. THE COMPOUND PENDULUM.
WORK, POWER AND IMPULSE
9.68. A constant force of magnitude F is applied tangentially to a flywheel which can rotate about a
fixed axis perpendicular to it and passing through its center. If the flywheel has radius a, radius
of gyration K and mass M, prove that the angular acceleration is given by F a/MK 2 .
9.69. How long will it be before the flywheel of Problem 9.68 reaches an angular speed w if it starts
from rest? Ans. MK 2 a /F a
9.70. Assuming that the flywheel of Problem 9.68 starts from rest, find (a) the total work done,
(6) the total power developed and (c) the total impulse applied in getting the angular speed
up to <o . Ans  ( a ) %MK 2 <»%> ( b ) *>«o. (c) MK 2 u
CHAP. 9]
PLANE MOTION OF RIGID BODIES
247
9.71. Work (a) Problem 9.68, (b) Problem 9.69 and (c) Problem 9.70 if F = 10 newtons, a = 1 meter,
K  0.5 meter, M = 20 kilograms and w = 20 radians/sec.
Ans. (a) 2 rad/sec 2 ; (b) 10 sec; (c) 250 joules, 200 joules/sec, 100 newton sec
9.72. Find the period of small vibrations for a simple pendulum assuming that the string supporting
the bob is replaced by a uniform rod of length I and mass M while the bob has mass m.
9.73.
9.74.
Ans. 2v a\
2(M + 3m)Z
3(M + 2m)g
Discuss the cases (a) M = and (b) m = in Problem 9.72.
A rectangular plate having edges of lengths a and b respectively hangs vertically from the edge
of length a. (a) Find the pe riod fo r small oscillations and (6) the length of the equivalent simple
pendulum. Ans. (a) 2wy/2b/Zg , (b) 6
9.75. A uniform solid sphere of radius a and mass M is suspended vertically downward from a point
on its surface, (a) Find the period for small oscillations in a plane and (6) the length of the
equivalent simple pendulum. Ans. (a) 2ir^la/hg, (b) 7a/5
9.76.
9.77.
A yoyo consists of a cylinder of mass 80 gm around which a string of length 60 cm is wound.
If the end of the string is kept fixed and the yoyo is allowed to fall vertically starting from rest,
find its speed when it reaches the end of the string. Ans. 280 cm/sec
Find the tension in the string of Problem 9.76.
Ans. 19,600 dynes
9.78. A hollow cylindrical disk of mass M moving with constant speed v
comes to an incline of angle a. Prove that if there is no slipping
it will rise a distance v\/{g sin a) up the incline.
9.79. If the hollow disk of Problem 9.78 is replaced by a solid disk, how
high will it rise up the incline? Ans. Svl/(4g sin a)
9.80. In Fig. 936 the pulley, assumed frictionless, has radius 0.2 meter
and its radius of gyration is 0.1 meter. What is the acceleration of
the 5 kg mass? Ans. 2.45 m/sec 2
Y/////////////////////'
20 kg
10 kg  
5 kg
Fig. 936
INSTANTANEOUS CENTER. SPACE AND BODY CENTRODES
9.81. A ladder of length I moves so that one end is on a vertical wall and the other on a horizontal
floor. Find (a) the space centrode and (b) the body centrode.
Ans. (a) A circle having radius I and center at point O where the floor and wall meet.
(b) A circle with the ladder as diameter
9.82.
9.83.
9.84.
A long rod AB moves so that it remains in contact with
the top of a post of height h while its foot B moves on a
horizontal line CD [Fig. 937]. Assuming the motion
to be in one plane, find the locus of instantaneous centers.
What is the (a) body centrode and (b) space centrode
in Problem 9.82?
Work Problems 9.82 and 9.83 if the post is replaced
by a fixed cylinder of radius a.
Fig. 937
STATICS OF A RIGID BODY
9.85. A uniform ladder of weight W and length I has its top against a smooth wall and its foot on a
floor having coefficient of friction p. (a) Find the smallest angle a which the ladder can make
with the horizontal and still be in equilibrium. (6) Can equilibrium occur if ft  0? Explain.
248
PLANE MOTION OF RIGID BODIES
[CHAP. 9
9.86. Work Problem 9.85 if the wall has coefficient of friction n v
9.87. In Fig. 938, AB is a uniform bar of length I and weight W supported at C. It carries weights W x
at A and W 2 at D so that AC — a and CD — b. Where must a weight W 3 be placed on AC
so that the system will be in equilibrium?
w 3
□ c
A
Fig. 938
W 2
D
Fig. 939
9.88. A uniform triangular thin plate hangs from a fixed point O by strings OA, OB and OC of
lengths a, b and c respectively. Prove that the tensions T it T % and T 3 in the strings are such that
T x /a = T 2 /b = T 3 /c.
9.89. A uniform plank AB of length I and weight W is supported
at points C and D distant a from A and b from B respec
tively [Fig. 939]. Determine the reaction forces at C and D
respectively.
9.90. In Fig. 940, OA and OB are uniform rods having the same
density and connected at O so that AOB is a right angle.
The system is supported at O so that AOB is in a vertical
plane. Find the angles a and /? for which equilibrium occurs.
Ans. a = tan 1 (a/b), /? = v/2 — tan 1 (a/b)
Fig. 940
MISCELLANEOUS PROBLEMS
9.91. A circular cylinder has radius a and height h. Prove that the moment of inertia about an axis
perpendicular to the axis of the cylinder and passing through the centroid is J^M(h 2 + 3a 2 ).
9.92. Prove that the effect of a force on a rigid body is not changed by shifting the force along its line
of action.
9.93. A cylinder of radius a and radius of gyration K rolls without slipping down an inclined plane of
angle a and length I, starting from res t at the top of the in cline. Prove that when it reaches the
bottom of the incline is speed will be ^(2gla 2 sin a)l(a 2 + K 2 ) .
9.94. A cylinder resting on top of a fixed cylinder is given a slight displacement so that it rolls without
slipping. Determine where it leaves the fixed cylinder.
Ans. — sin 1 4/7 where e is measured as in Fig. 933, page 244.
9.95. Work Problem 9.42 if the sphere is given an initial speed v .
9.96. Work Problem 9.94 if the cylinder is given an initial speed v .
9.97. A sphere of radius a and radius of gyration K about a diameter rolls without slipping down
an incline of angle a. Prove that it descends with constant acceleration given by (ga 2 sin a)/(a 2 + K 2 ).
9.98. Work Problem 9.97 if the sphere is (a) solid, (b) hollow and of negligible thickness.
Ans. (a) f g sin a, (&) f g sin a
9.99. A hollow sphere has inner radius a and outer radius 6. Prove that if M is its mass, then the
moment of inertia about an axis through its center is
'a 4 + a 3 & + a 2 & 2 + a& 3 + & 4>
M
i 2 + ab + &2
Discuss the cases 6 = and a — b.
CHAP. 9]
PLANE MOTION OF RIGID BODIES
249
9.100. Wooden plates, all having the same rectangular
shape are stacked one above the other as indicated
in Fig. 941. (a) If the length of each plate is
2a, prove that equilibrium conditions will prevail
if the (n + l)th plate extends a maximum distance
of a/n beyond the wth plate where n = 1,2,3, ... .
(b) What is the maximum horizontal distance
which can be reached if more and more plates
are added?
9.101. Work Problem 9.100 if the plates are stacked on
a sphere of radius R instead of on a flat surface
as assumed in that problem.
9.107.
Fig. 941
9.102. A cylinder of radius a rolls on the inner surface of a smooth cylinder of radius 2a. Prove that
the period of small oscillations is 2v 1 \/3a/2g .
9.103. A ladder of length I and negligible weight rests with one end against a wall having coefficient of
friction ^ and the other end against a floor having coefficient of friction /i 2 . It makes an angle a
with the floor, (a) How far up the ladder can a man climb before the ladder slips? (6) What is
the condition that the ladder not slip at all regardless of where the man is located?
Ans. (a) /j. 2 l(m + t&n a)/ (finx 2 + 1), (b) tana > l//t 2
9.104. Work Problem 9.103 if the weight of the ladder is not
negligible.
9.105. A ladder AB of length I [Fig. 942] has one end A on an
incline of angle a and the other end B on a vertical wall.
The ladder is at rest and makes an angle /? with the incline.
If the wall is smooth and the incline has coefficient of
friction n, find the smallest value of n so that a man of
weight W m will be able to climb the ladder without having it
slip. Check your answer by obtaining the result of Problem
9.37, page 241, as a special case.
9.106. Work Problem 9.105 if the wall has coefficient of friction
0i
Fig. 942
A uniform rod AB with point A fixed rotates about a vertical axis so that it makes a constant
angle a with the vertical [Fig. 943], If th e length of the rod is I, prove that the angular
speed needed to do this is « = ^J(3g sec a)/2l .
'//////////////////'.
9.108.
Fig. 943
Fig. 944
Fig. 945
A circular cylinder of mass m and radius a is suspended from the ceiling by a wire as shown in
Fig. 944. The cylinder is given an angular twist 6 and is then released. If the torque is assumed
proportional to the angle through which the cylinder is turned and the constant of proportionality
is X, prove that the cylinder will undergo simple harmonic motion with period 2iray/m/2\ .
9.109. Find the period in Problem 9.108 if the cylinder is replaced by a sphere of radius a.
Ans. 2jra\/2m/5X
9.110. Work (a) Problem 9.108 and (6) Problem 9.109 if damping proportional to the instantaneous
angular velocity is present. Discuss physically.
9.111. A uniform beam AB of length I and weight W [Fig. 945] is supported by ropes AC and BD of
lengths a and 6 respectively making angles a and /? with the ceiling CD to which the ropes
are fixed. If equilibrium conditions prevail, find the tensions in the ropes.
250
PLANE MOTION OF RIGID BODIES
[CHAP. 9
9.112. In Fig. 946 the mass m is attached to a rope which is wound around
a fixed pulley of mass M and radius of gyration K which can rotate
freely about O. If the mass is released from rest, find (a) the angular
speed of the pulley after time t and (b) the tension in the rope.
9.113. Prove that the acceleration of the mass m in Problem 9.112 is
ga 2 /(a 2 + K 2 ).
9.114. Describe how Problem 9.112 can be used to determine the radius of
gyration of a pulley.
w////////////////.
9.115.
9.116.
9.117.
9.120.
9.123.
Fig. 946
A uniform rod AB [Fig. 947] of length I and weight W having its
ends on a frictionless wall OA and floor OB respectively, slides
starting from rest when its foot B is at a distance d from O. Prove
that the other end A will l eave the wall when the foot B is at a
distance from O given by ^yoT^ + 4d 2 .
A cylinder of mass 10 lb rotates about a fixed horizontal axis through
its center and perpendicular to it. A rope wound around it carries
a mass of 20 lb. Assuming that the mass starts from rest, find its
speed after 5 seconds. Ans. 128 ft/sec
Fig. 947
What must be the length of a rod suspended from one end so that it will be a seconds pendulum
on making small vibrations in a plane? Ans. 149 cm
9.118. A solid sphere and a hollow sphere of the same radius both start from rest at the top of an
inclined plane of angle a and roll without slipping down the incline. Which one gets to the
bottom first? Explain. Ans. The solid sphere
9.119. A compound pendulum of mass M and radius of gyration K about a horizontal axis is displaced
so that it makes an angle O with the vertical and is then released. Prove that if the center of
mass is at distance a from the axis, then the reaction force on the axis is given by
Mg
#2 + a 2
y/([K 2 + 2a 2 ] cos*  a 2 cos ) 2 + (K 2 sin 0) 2
A rectangular parallelepiped of sides a, b, and c is suspended vertically from the side of length a.
Find the period of small oscillations.
9.121. Find the least coefficient of friction needed to prevent the sliding of a circular hoop down an
incline of angle a. Ans. § tan a
9.122. Find the period of small vibrations of a rod of length I suspended vertically about a point %l
from one end.
9.124.
9.125.
A pulley system consists of two solid disks of radius r x and r 2
respectively rigidly attached to each other and capable of rotating
freely about a fixed horizontal axis through the center O. A weight
W is suspended from a string wound around the smaller disk as shown
in Fig. 948. If the radius of gyration of the pulley system is K
and its weight is w, find (a) the angular acceleration with which the
weight descends and (b) the tension in the string.
Ans. (a) WgrJiWri + wK 2 ), (b) WwK 2 /(Wr\ + wK 2 )
A solid sphere of radius 6 rolls on the inside of a smooth hollow
sphere of radius a. Prove that the period for small oscillations is
given by 2vy/7(a — b)/5g .
A thin circular solid plate of radius a is suspended vertically from a
horizontal axis passing through a chord AB [see Fig. 949]. If it
makes small oscillations about this axis, prove that the frequency
of such oscillations is greatest when AB is at distance a/2 from
the center.
Fig. 949
CHAP. 9]
PLANE MOTION OF KIGID BODIES
251
9.126. A uniform rod of length 51 is suspended vertically from a string of length 21 which has its other
end fixed. Prove that the normal frequencies for small oscillations in a plane are — \\— and
1 [^ 2 ^ * 5l
1_
2ir
y , and describe the normal modes.
9.127. A uniform rod of mass m and length I is suspended from one of its ends. What is the minimum
speed with which the other end should be hit so that it will describe a complete vertical circle?
9.128.
9.129.
(a) If the bob of a simple pendulum is a uniform solid sphere of radius a rather than a point
mass, prove that the period for small oscillations is 2iry/l/g + 2a 2 /5gl .
(b) For what value of I is the period in (a) a minimum?
A sphere of radius a and mass M rolls along a horizontal plane with constant speed v . It comes
to an incline of angle a. Assuming that it rolls without slipping, how far up the incline will it
travel? Ans. 10v§/(7flr sin a)
9.130. Prove that the doughnut shaped solid or torus of Fig. 950 has
a moment of inertia about its axis given by Af(3a 2 + 4b 2 ).
9.131. A cylinder of mass m and radius a rolls without slipping down
a 45° inclined plane of mass M which is on a horizontal friction
less table. Prove that while the rolling takes place the incline
will move with an acceleration given by mg/(SM + 2m).
9.132. Work Problem 9.131 if the incline is of angle a.
Ans. (mg sin 2a)/(3M + 2m  ra cos 2a)
9.133. Find the (a) tension in the rope and (6) accelera
tion of the system shown in Fig. 951 if the radius
of gyration of the pulley is 0.5 m and its mass
is 20 kg.
Fig. 950
200 kg
n = .2
1\
9.134. Compare the result of Problem 9.133 with that
obtained assuming the pulley to have negligible mass.
9.135. Prove that if the net external torque about an axis
is zero, then it is also zero about any other axis.
a
100 kg
Fig. 951
9.136. A solid cylindrical disk of radius a has a circular hole of radius b whose center is at distance c
from the center of the disk. If the disk rolls down an inclined plane of angle a, find its acceleration.
[See Fig. 952/
y
Fig. 952
Fig. 953
9.137. Find the moment of inertia of the region bounded by the lemniscate r 2 = a 2 cos 2e [see Fig 9531
about the x axis. Ans. Ma 2 (3ir  8)/48
9.138. Find the largest angle of an inclined plane down which a solid cylinder will roll without slipping
if the coefficient of friction is /t.
9.139. Work Problem 9.138 for a solid sphere.
252
PLANE MOTION OF RIGID BODIES
[CHAP. 9
9.140. Discuss the motion of a hollow cylinder of inner radius a and outer radius b as it rolls down an
inclined plane of angle a.
9.141. A table top of negligible weight has the form of an equilateral triangle ABC of side s. The legs
of the table are perpendicular to the table top at the vertices. A heavy weight W is placed on
the table top at a point which is distant a from side BC and 6 from side AC. Find that part
of the weight supported by the legs at A, B and C respectively.
Ans.
2Wa 2Wb
Wll
2a + 26
sV3
9.142. Discuss the motion of the disk of Problem 9.136 down the inclined plane if the coefficient of
friction is /<.
9.143. A hill has a cross section in the form of a cycloid
x = a{e + sin $), y — a(l — cos 9)
as indicated in Fig. 954. A solid sphere of
radius b starting from rest at the top of the hill
is given a slight displacement so that it rolls
without slipping down the hill. Find the speed
of its center when it reaches the bottom of the hill.
Ans. y/l0g(2a  b)/l
Fig. 954
9.144. Work Problem 3.108, page 85, if the masses and moments of inertia of the pulleys are taken
into account.
9.145. Work Problem 9.38, page 242, if friction is taken into account.
9.146. A uniform rod of length I is placed upright on a table and then allowed to fall. Assuming that its
point of contact with the table does not move, prove that its angula r velocity at the instant when
it makes an angle d with the vertical is given in magnitude by y/3g{icose)/2l .
9.147. Prove Theorem 9.11, page 228, for the case where the vibrations are not necessarily small.
Compare Problem 9.41, page 243.
9.148. A rigid body moves parallel to a given fixed plane. Prove that there is one and only one point
of the rigid body where the instantaneous acceleration is zero.
9.149. A solid hemisphere of radius a rests with its convex surface on a horizontal table. If it is
displaced slightly, prove that it will undergo oscillations with period equal to that of a simple
pendulum of equivalent length 4a/3.
9.150. A solid cylinder of radius a and height h is suspended from axis AB
as indicated in Fig. 955. Find the period of small oscillations about A
this axis.
9.151. Prove that a solid sphere will roll without slipping down an inclined
plane of angle a if the coefficient of friction is at least f tan a.
9.152. Find the least coefficient of friction for an inclined plane of angle a
in order that a solid cylinder will roll down it without slipping.
Ans. ^ tan a
~«e*%%$mk
B
Fig. 955
Chapter 10
SPACE MOTION
of RIGID BODIES
GENERAL MOTION OF RIGID BODIES IN SPACE
In Chapter 9 we specialized the motion of rigid bodies to one of translation of the
center of mass plus rotation about an axis through the center of mass and perpendicular
to a fixed plane. In this chapter we treat the general motion of a rigid body in space.
Such general motion is composed of a translation of a fixed point of the body [usually the
center of mass] plus rotation about an axis through the fixed point which is not necessarily
restricted in direction.
DEGREES OF FREEDOM
The number of degrees of freedom [see page 165] for the general motion of a rigid body
in space is 6, i.e. 6 coordinates are needed to specify the motion. We usually choose 3 of
these to be the coordinates of a point in the body [usually the center of mass] and the
remaining 3 to be angles [for example, the Euler angles, page 257] which describe the
rotation of the rigid body about the point.
If a rigid body is constrained in any way, as for example by keeping one point fixed,
the number of degrees of freedom is of course reduced accordingly.
PURE ROTATION OF RIGID BODIES
Since the general motion of a rigid body can also be expressed in terms of translation
of a fixed point of the rigid body plus rotation of the rigid body about an axis through the
point, it is natural for us to consider first the case of pure rotation and later to add the
effects of translation. To do this we shall first assume that one point of the rigid body
is fixed in space. The effects of translation are relatively easy to handle and can be obtained
by using the result (10), page 167.
VELOCITY AND ANGULAR VELOCITY OF
A RIGID BODY WITH ONE POINT FIXED
Suppose that point O of the rigid body % of Fig. 101
is fixed. Then at a given instant of time the body
will be rotating with angular velocity <■> about the in
stantaneous axis through O. A particle P of the body
having position vector r„ with respect to O will have
an instantaneous velocity v v given by
— r v — »Xr»
CO
See Problem 10.2.
Fig. 101
253
254 SPACE MOTION OF RIGID BODIES [CHAP. 10
ANGULAR MOMENTUM
The angular momentum of a rigid body with one point fixed about the instantaneous
axis through the fixed point is given by
= 2 ™< v {t v Xx v ) = 2 ™<v{rv X («» X r v )} {2)
where m v is the mass of the vth particle and where the summation is taken over all
particles of %.
MOMENTS OF INERTIA. PRODUCTS OF INERTIA
Let us choose a fixed xyz coordinate system having origin O and let us write
a = a£ + a y j + nji, « = <*j. + o> y j + ajs.
r v = Xvi + y v j + z v k
Then equation (2) can be written in component form as [see Problem 10.3].
Q x = 7 *z w s + 7 xy<% + I xz"z
9 y = ty^x + hy% + hz^z
Q z ~ hx^x + hy% + hz^z
(•4)
(6)
where l xx = ^m^ + z*), I yy = Jm^ + 4 I zz = ^m^ + yl) (5)
Ixy — ^ Wl v X v yv — ':
lyz   2 m " V* z »  I
lzx — — • ■ W/v Z11 Xv — lxz
The quantities hx, Iyy, Lz are called the moments of inertia about the x, y and z axes respec
tively. The quantities I xy , hz, ... are called products of inertia. For continuous mass
distributions these can be computed by using integration.
Note that the products of inertia in (6) have been defined with an associated minus
sign. As a consequence minus signs are avoided in (4).
MOMENT OF INERTIA MATRIX OR TENSOR
The nine quantities I xx , I xy , ...,hz can be written in an array often called a matrix or
tensor given by
lxx lxy la
lyx Iyy lyz I V /
\lzx Izy lz
and each quantity is called an element of the matrix or tensor. The diagonal consisting
of the elements hxjyyjzz is called the principal or main diagonal. Since
Ixy — lyx, Ixz — lzx, lyz = Izy \P)
it is seen that the elements have symmetry about the main diagonal. For this reason (7)
is often referred to as a symmetric matrix or tensor.
KINETIC ENERGY OF ROTATION
The kinetic energy of rotation is given by
T = Wxxl + hy< + hzl + 2I xy<»x<»y + ^ xz^z + 2I yz<»y<»z)
= i<o0 {9)
CHAP. 10]
SPACE MOTION OF RIGID BODIES
255
PRINCIPAL AXES OF INERTIA
A set of 3 mutually perpendicular axes having origin O which are fixed in the body
and rotating with it and which are such that the products of inertia about them are zero,
are called principal axes of inertia or briefly principal axes of the body.
An important property of a principal axis [which can also be taken as a definition]
is that if a rigid body rotates about it the direction of the angular momentum is the same
as that of the angular velocity. Thus
O = Jo, (10)
where / is a scalar. From this we find [see Problem 10.6] that
Vx + ^^K + U = ° !► («)
l zx<»x + l zy (A y + Vzz  7 )°
=
In order that (11) have solutions other than the trivial one w x = 0, a y = 0, » z = 0, we
require that
ixx 1 Ixy *xz
lyx IyyI hz = <> (12)
Izx izy Izz L
This leads to a cubic equation in / having 3 real roots h,h,h. These are called the
principal moments of inertia. The directions of the principal axes can be found from (11),
as shown in Problem 10.6 by finding the ratio o» x : w : «> 2 .
An axis of symmetry of a rigid body will always be a principal axis.
ANGULAR MOMENTUM AND KINETIC ENERGY ABOUT
THE PRINCIPAL AXES
If we call <a v <a 2 , (o 3 and Q lt Q 2 , o 3 the magnitudes of the angular velocities and angular
momenta about the principal axes respectively, then
°i — A* !* n 2 _7 2 a, 2' n 3  h^s
The kinetic energy of rotation about the principal axes is given by
t = WA + *A + W)
which can be written in vector form as [compare equation (9)]
T = a>0
(13)
(U)
(15)
THE ELLIPSOID OF INERTIA
Let n be a unit vector in the direction of <■>. Then
o> = a>n = w(C0S a i + COS P j + COS y k)
(16)
where cos a, cos /3, cos y are the direction cosines of « or n with respect to the x, y and z
axes. Then the kinetic energy of rotation is given by
T = \h
(17)
256
SPACE MOTION OF RIGID BODIES
[CHAP. 10
where I = I xx cos 2 a + I yy cos 2 ft + I zz cos 2 y
+ 27 xy COS a COS /? + 27 ' COS /3 COS y + 27 xz COS a COS y
By defining a vector
p = n/yT
where p = Px i + P J + P k, (18) becomes
7 p 2 + 7 p 2 + 7 p 2 + 27 P p+ 27 P p+ 27 op =1
xxrx yy"y zz"z xyrxry yzryrz zx"z"x
(18)
(19)
(20)
In the coordinates p x ,p y ,p z equation (20) represents an ellipsoid which is called the
ellipsoid of inertia or the momental ellipsoid.
If the coordinate axes are rotated to coincide with the principal axes of the ellipsoid,
the equation becomes
*A + h?\ + 1 A = 1 ( 21 )
where p v p 2 ,p 3 represent the coordinates of the new axes.
EULER'S EQUATIONS OF MOTION
It is convenient to describe the motion of a rigid body relative to a set of coordinate
axes coinciding with the principal axes which are fixed in the body and thus rotate as the
body rotates. If A lf A 2 , a 3 and c^, <o 2 , w 3 represent the respective components of the external
torque and angular velocity along the principal axes, the equations of motion are given by
Vl + (*s _ ^K^ = A l
h^2 + (*1 ~ ^sK^l ~ A 2
Vs + (^2 _ ^l) *^ ~ A 3
These are often called Euler's equations.
(22)
FORCE FREE MOTION.
THE INVARIABLE LINE AND PLANE
Suppose that a rigid body is rotating about a
fixed point O and that there are no forces acting
on the body [except of course the reaction at the
fixed point]. Then the total external torque is
zero. Thus the angular momentum vector Q is
constant and so has a fixed direction in space as
indicated in Fig. 102. The line indicating this
direction is called the invariable line.
Since the kinetic energy is constant [see
Problem 10.34], we have from (15)
wo = constant
(23)
Fig. 102
This means that the projection of » on O is constant, so that the terminal point of <*
describes a plane. This plane is called the invariable plane.
As the rigid body rotates, an observer fixed relative to the body coordinate axes would
see a rotation or precession of the angular velocity vector a> about the angular momentum
vector CI.
CHAP. 10]
SPACE MOTION OF RIGID BODIES
257
POINSOTS CONSTRUCTION.
SPACE AND BODY CONES
As noted by Poinsot, the above ideas can
be geometrically interpreted as a rolling
without slipping of the ellipsoid of inertia
corresponding to the rigid body on the
invariable plane. The curve described on the
invariable plane by the point of contact with
the ellipsoid is called the herpolhode [see
Fig. 103]. The corresponding curve on the
ellipsoid is called the polhode.
To an observer fixed in space it would
appear that the vector a traces out a cone
which is called the space cone. To an
observer fixed on the rigid body it would
appear that <■> also traces out a cone which is
called the body cone. The motion can then
be equivalently described as a rolling without
slipping of one cone on the other. See
Problem 10.19.
POLHODE. HERPOLHODE.
Fig. 103
SYMMETRIC RIGID BODIES. ROTATION OF THE EARTH
Simplifications occur in the case of a symmetric rigid body. In such case at least
two principal moments of inertia, say h and h, are equal and the ellipsoid of inertia
is an ellipsoid of revolution. We can then show [see Problem 10.17] that the angular
velocity vector <» precesses about the angular momentum vector O with frequency given by
/
1
hh
2tt
h
in)
where the constant A is the component of the angular velocity in the direction of the
axis of symmetry.
In the case of the earth, which can be assumed to be an ellipsoid of revolution flattened
slightly at the poles, this leads to a predicted precession period of about 300 days. In
practice, however, the period is found to be about 430 days. The difference is explained as
due to the fact that the earth is not perfectly rigid.
THE EULER ANGLES
In order to describe the rotation of a rigid
body about a point we use 3 angular coordinates
called Euler angles. These coordinates denoted
by 4>, 9, xp are indicated in Fig. 104. In this
figure the xyz coordinate system can be rotated
into the x'y'z' system by successive rotations
through the angles <£ and then 6 and then ^
[see Problem 10.20]. The line OA is sometimes
called the line of nodes.
In practice the x',y',z' axes are chosen as
the principal axes or body axes of the rigid
body while the x, y and z axes or space axes
are fixed in space.
Fig. 104
258
SPACE MOTION OF RIGID BODIES
[CHAP. 10
ANGULAR VELOCITY AND KINETIC ENERGY
IN TERMS OF EULER ANGLES
w l» w 2' w 3
In terms of the Euler angles the components
x f , y' and z' axes are given by
<£ sin sin ^ + 6 sin ip
$ sin 6 cos xp — 6 sin ^
<£ cos 8 + }
The kinetic energy of rotation is then given by
T = *(/,«? + V5 + W)
where h, h, h are the principal moments of inertia.
of the angular velocity along the
(25)
(26)
MOTION OF A SPINNING TOP
An interesting example of rigid body motion
occurs when a symmetrical rigid body having one
point on the symmetry axis fixed in space is set
spinning in a gravitational field. One such example
is that of a child's top as shown in Fig. 105, where
point O is assumed as the fixed point.
For a discussion of the various kinds of motion
which can occur, see Problems 10.2510.32 and 10.36.
Fig. 105
GYROSCOPES
Suppose a circular disk having its axis mounted in gimbals [see Fig. 106] is given a
spin of angular velocity a. If the outer gimbal is turned through an angle, the spin axis
of the disk will tend to point in the same direction as previously [see Fig. 107]. This
assumes of course that friction at the gimbal bearings is negligible.
In general the direction of the spin axis remains fixed even when the outer gimbal,
which is attached to some object, moves freely in space. Because of this property the
mechanism, which is called a gyroscope, finds many applications in cases where maintaining
direction [or following some specified course] is important, as for example in navigation
and guidance or control of ships, airplanes, submarines, missiles, satellites or other moving
vehicles.
A gyroscope is another example of a symmetric spinning rigid body with one point on
the symmetry axis [usually the center of mass] taken as fixed.
Spin
Fig. 106
Fig. 107
CHAP. 10] SPACE MOTION OF RIGID BODIES 259
Solved Problems
GENERAL MOTION OF RIGID BODIES IN SPACE
10.1. Find the number of degrees of freedom for a rigid body which (a) can move freely
in space, (b) has one point fixed, (c) has two points fixed.
(a) 6 [see Problem 7.2(a), page 172]
(6) 3 [see Problem 7.2(6), page 172]
(c) If two points are fixed, then the rigid body rotates about the axis joining the two fixed
points. Then the number of degrees of freedom is 1, such as for example the angle of rotation
of the rigid body about this axis.
10.2. A rigid body undergoes a rotation of angular velocity a about a fixed point O. Prove
that the velocity v of any particle of the body having position vector r relative to O
is given by v = «xr.
This follows at once from Problem 6.1, page 147, on noting that the velocity relative to the
moving system is dr/dt\ M = dr/dt\ b = 0.
ANGULAR MOMENTUM. KINETIC ENERGY.
MOMENTS AND PRODUCTS OF INERTIA
10.3. Derive the equations (4), page 254, for the components of angular momentum in terms
of the moments and products of inertia given by equations (5) and (6), page 254.
The total angular momentum is given by
N N
G = 2 m„(r„ XvJ = 2 m v {r v X ( w X r„)}
v=l v=l
where we have used Problem 10.2 applied to the ?th particle.
Now by equation (7), page 5, we have
r„ X (« X r„) = «(r„ • r„) — r„(«* • r„)
= (w x i + u„j + w x k)(*2 + y% + z v )
 {x v i + y v \ + z v k)(a x x v + Uy y v + u> z z v )
 WxiVv + z l)  UyXvVv  <*z x V Z V }i
+ {a y («$ + z\)  UxXvVv ~ u z Vv z v}J
+ {o> z (xl + y v ) — u x x v z v — u y y v z v }k
Then multiplying by m„, summing over v and equating the coefficients of i, j and k to Sl x , S2 y and Sl z
respectively, we find as required
«x = ]2 ™ v (yl +z*)><* x + < — 2 m v x v y v > «„ + < — 2 m v x v z v
— Ixx u x "f" •'xj/ w j/ + *xz u z
Vy = ] 2 m v X v y v > OJ x + j 2 ™>v(xl + *v) f «y + \ ~ 2 ™>vVvZv \ ">x
= Ixy u x + lyy^y ~^~ lyz^z
 2 nh*v*v f u x + "j ~ 2 rn^y^, > a y + l 2 ™>v(tf + Vv)\<*z
= *xz u x + lyz^y + *zz u z
For continuous mass distributions of density a, we can obtain the same results by starting with
O =  a(rX\)dr =  <x{r X (<• X r)} dr
260
SPACE MOTION OF RIGID BODIES
[CHAP. 10
10.4. If a rigid body with one point fixed rotates with angular velocity w and has angular
momentum CI, prove that the kinetic energy is given by T = <*> • CI.
= 2m,{(»Xr,)(«Xr v )} = £ 2 w„{« • [r„ X (. X r„)]}
= £• • 2 wi„r„ X((.Xr„) = i tt 0
N
where we have used the abbreviation 2 in place of 2 •
10.5. Prove that the kinetic energy in Problem 10.4 can be written
T = W X A + I y y y + I Z A + 21 xM + 2I xz ^ z + 2I yz ^ z )
From Problem 10.4, we have
T = l«> • CI
i{«a;(4a; w x + 4i/ w y + 4z"«)
+ Uyilyx^x + ^«/y w y + ^2/Z w z)
+ w z(4x w a: + fz2,«3, + 4z«z)}
i(^xx w x + lyyUy + I Z z<*z + ^xy u x u y + ^xz a x u z + 2I yz w y a z )
using
the fact that /«. = I vx , I xz = I zx , I.
/„..
PRINCIPAL MOMENTS OF INERTIA AND PRINCIPAL AXES
10.6. Derive equations (11), page 255, for the principal moments of inertia and the
directions of the principal axes.
Using CI = I*
together with equations (3) and (4), page 254, we have
(1)
= Ia y
xz u x ~r *uz u u + 'z2« z — "*
"■xy^y
*yx u x + ^j/y w r/ + ^j/« w x:
l yz>»y
(hx ~ I)<*x + ^"j/ + 4« w z =
lyx^x + Uj/j/ ~ I) u y "+" ^j/z w z =
4z«x + lyz^y + (I zz ~ I)<*z — °
(2)
The principal moments of inertia are found by setting the determinant of the coefficients of
u x , o) y , u z in (2) equal to zero, i.e.,
* T.V. *
*xy
*yy ~ *
'■xz
*yz
I,, I
=
This is a cubic equation in / leading to three values I lt / 2> I3 which are the principal moments
of inertia. By putting I = 7 X in (2) we obtain ratios for a x : u y : u z which yields the direction of »
or the direction of the principal axis corresponding to I v Similarly, by substituting I 2 and I 3
we find the directions of the corresponding principal axes.
CHAP. 10]
SPACE MOTION OF RIGID BODIES
261
10.7. Find the (a) moments of inertia and (b) products of inertia of a uniform square
plate of length a about the x, y and z axes chosen as shown in Fig. 108.
(a) The moment of inertia of an element dx dy of the plate
about the x axis if the density is a is ay 2 dx dy. Then the
moment of inertia of the entire plate about the x axis is jjj
/a pa
I ay 2 dx dy
a»4 = L M a 2 (1)
since the mass of the plate is M — aa 2 .
Similarly, the moment of inertia of the plate about
the y axis is
J id pa
I a x 2 dxdy = i<ra 4 = ±Ma 2 (2)
as is also evident by symmetry.
f— u—  — dx dy
Fig. 108
The moment of inertia of dx dy about the z axis is a{x 2 + y 2 ) dx dy, and so the moment of
inertia of the entire plate about the z axis is
hz = f" f <r(x 2 + y 2 ) dx dy = %Ma 2 + %Ma 2 = \Ma 2 (3)
This also follows from the perpendicular axes theorem [see page 226].
(6) The product of inertia of the element dx dy of the plate about the x and y axes is axydxdy,
and so the product of inertia of the entire plate about these axes is
= /,.
pa pa
— — I I axy dx dy — — j<t« 4
\Ma 2
4
(4)
x=0 *'a =
The product of inertia of the element dx dy of the plate about the * and z axes is the
product of adxdy by the distances to the yz and xy planes, which are x and respectively.
Thus we must have
= /,
0,
and similarly
hu =
(5)
%Ma 2  I
\Ma 2
\Ma 2
\Ma 2  I
Ma 2 
 J
10.8. Find the (a) principal moments of inertia and (b) the directions of the principal
axes for the plate of Problem 10.7.
(a) By Problem 10.6 and the results (l)(5) of Problem 10.7, we obtain
(1)
or [(%Ma 2  I)(±Ma 2  I)  (lMa 2 )(lMa 2 )][%Ma 2  I] =
which can be written
[P  Ma2J + £M 2 a*][%Ma 2  1} =
Setting the first factor equal to zero and using the quadratic formula to solve for /, we
find for the three roots of (1),
I^^Ma 2 , / 2 = JLMa2, 7 3 = \Ma 2 (2)
which are the principal moments of inertia.
(6) To find the direction of the principal axis corresponding to I lt we let I = I x = j^Ma 2 in
the equations
(±Ma 2 I)a x  lMa 2 u y + <o 2 =
±Ma 2 a> x + (±Ma 2 I)a y + <* z = [ (3)
0<a x + Wy + (Ma27)a> 2 =
262
SPACE MOTION OF RIGID BODIES
[CHAP. 10
The first two equations yield u y — w x while the third gives w x = 0. Thus the direction
of the principal axis is the same as the direction of the angular velocity vector
<■» = w x i + w y j + u z k — u x i + uj = u x (i + j)
Then the principal axis corresponding to /j is in the direction i + j.
Similarly, by letting I — I 2 ~ ^Ma 2 in (3) we find
u y — —u x , u z = so that the direction of the correspond
ing principal axis is » = u x i — u x j = a x (i — j) or i — j.
If we let I = I 3 = %Ma 2 in (3) we find a x = 0,
u y = while u z is arbitrary. This gives <■> = to z k which
shows that the third principal axis is in the direction k.
The directions of the principal axes are indicated by
i + j, i — j and k in Fig. 109. Note that these are
mutually perpendicular and that i + j and i — j have the
directions of the diagonals of the square plate which are
lines of symmetry.
The principal moments of inertia can also be deter
mined by recognizing the lines of symmetry. Fig. 109
10.9. Find the principal moments of inertia at the center of a uniform rectangular plate
of sides a and b.
The principal axes lie along the directions of symmetry and thus must be along the x axis,
y axis and z axis [the last of which is perpendicular to the xy plane] as in Fig. 1010.
By Problems 9.6, 9.9 and 9.11 the principal moments of inertia are found to be I t = ^Ma 2 ,
I 2 = ^Mb 2 , I 3 = ^M(a 2 + b 2 ).
y
O
dr = dz dy dx
V* 2 + v 2
Fig. 1010
Fig. 1011
10.10. Find the principal moments of inertia at the center of the ellipsoid
,.2 „,2 ^2
= 1
— __ ^— 4 —
a"
One eighth of the ellipsoid is indicated in Fig. 1011. The moment of inertia of the volume
element dr of mass a dr about the z axis or "3" axis is (x 2 + y 2 )a dr, and the total moment of
inertia about the z axis is
yJl x 2 /a 2 s*c\ll  (r
r r x
2/„2 , .,2,^2,
(x 2 + y 2 )a dz dy dx
x =o "y=0
Integration with respect to z gives
pa s>b\/l  x 2 /a 2
8ac J J (* 2 + V 2 ) Vl  (aW + y 2 /b 2 ) dy dx
x=0 y=0
To perform this integration let x — aX, y — bY where X and Y are new variables. Then the
integral can be written
j (a 2 X 2 + b 2 Y 2 ) Vl  {X 2 + Y 2 ) dY dX
CHAP. 10]
SPACE MOTION OF RIGID BODIES
263
Introducing polar coordinates R, 6 in this XY plane, this becomes
• 1 s*tt/2
'R = *^0 = O
} (a 2 R 2 cos 2 + 6 2 « 2 sin 2 6) Vl  R 2 R dR dQ
= 2iraabc{cfi + b 2 )  R^VT^WdR = ±7roabc(a 2 + b 2 )
R =
where we use the substitution 1 — R 2 = t/ 2 in evaluating the last integral.
Since the volume of the ellipsoid is ^nabc, the mass is M = ^Traabc and hence 7 3
By symmetry we find /j = iM(6 2 + c 2 ), J 2 = iM(o 2 + c 2 ).
LM(a 2 + ft 2 ).
10.11. Suppose that the ellipsoid of Problem 10.10 is an oblate spheroid such that a = b
while c differs slightly from a or b. Prove that to a high degree of approximation,
(7 3  7i)//i = 1  c/a.
From Problem 10.10, if a — b then
a 2 + c 2
(a — c)(a + c)
a 2  c 2
But if c differs
only slightly from a then a + c « 2a and a 2 + c 2 « 2a 2 . Thus, approximately,
(/ 3  I x )/I = (a  c)(2a)/2a 2 = 1  c/a
10.12. Work Problem 10.11 for the case of the earth assumed to be an oblate spheroid.
Since the polar diameter or distance between north and south poles is very nearly 7900 miles
while the equatorial diameter is very nearly 7926 miles, then taking the polar axis as the "3" axis
we have 2c = 7900, 2a = 7926 or c = 3950, a = 3963.
Thus by Problem 10.11, (Z 3  h)/I x = 1 ~ 3950/3963 = .00328.
ELLIPSOID OF INERTIA
10.13. Suppose that the moments and products of inertia of a rigid body % with respect
to an xyz coordinate system intersecting at origin O are I xx , hv, Izz> I*y> Ixz, hz respec
tively. Prove that the moment of inertia of % about an axis making angles a, p, y
with the x, y and z axes respectively is given by
7 = Ixx COS 2 a + I yy COS 2 P + Izz COS 2 y
+ 21 xy COS a COS P + 21 xz COS a COS y + 21 yz COS p COS y
by
A unit vector in the direction of the axis is given
n = cos a i + cos )3 j + cos Y k
Then if m„ has position vector r„, its moment of inertia
about the axis OA is m v T)1 where D v  r„ X n. But
and
i J k
r„Xn = x v y v z v
cos a cos p cos y
= (y v cos y — z v cos /3)i
+ (z v cos a — x v cos y)j
+ (x v cos /3 — y v cos a)k Fig. 1012
\ Tv x n 2 = (y v cos y — z v cos (3) 2 + (z v cos a  x v cos y) 2 + (*„ cos fi  y v cos a) 2
= (vl + 4) cos 2 a + (x 2 + 4) cos 2 £ + (x 2 v + yl) cos 2 y
 2x v y v cos a cos (3  2x„z„ cos a cos y — 2y v z v cos /? cos y
264 SPACE MOTION OF RIGID BODIES [CHAP. 10
Thus the total moment of inertia of all masses ra„ is
= J2 ™v(vl + 4) I co S 2 a +  2 m^ajj! + 4) \ cos 2 /? + 1 2 m^ 2 . + yl) I cos 2 y
+ 2 j  2 myX v V v Y cos a cos + 2 j — 2 VA f cos « cos y
+ 2^2 m v y v z v > cos /? cos y
= I xx cos 2 a + lyy cos 2 /? + /^ cos 2 y
+ 2/j.j, cos a cos /? + 2/ X2 cos a cos y + 2I yz cos /? cos y
10.14. Find an equation for the ellipsoid of inertia corresponding to the square plate of
Problem 10.7.
We have from Problem 10.7,
I xx = \Ma\ l m = ±Ma 2 , I zz = *Ma*, I xy = iMa 2 , I xz = 0, /„, =
Then the equation of the ellipsoid of inertia is by equation (20), page 256,
\MaVx + iMaVy + \Mtfpl  \Ma? 9xPy = 1
or ?\ + P% + 2 P l  h x Py = 3/Ma*
EULER'S EQUATIONS OF MOTION
10.15. Find a relationship between the time rate of change of angular momentum of a rigid
body relative to axes fixed in space and in the body respectively.
If the rigid body axes are chosen as principal axes having directions of the unit vectors
e lt e 2 and e 3 respectively, then the angular momentum becomes
CI = ^U)^ + / 2 «2 e 2 + ^3"3C3
Now by Problem 6.1, page 147, if s and b refer to space (fixed) and body (moving) axes
respectively, then
at Is dt \b
= hu^! + 7 2 « 2 e 2 + h^3 e 3
+ ( W 1©1 + "2^2 + "3^3) X (/l"iei + I 2 ^2^2 + h w aPs)
— { Jl"l + (^3 ~ ■^2) w 2' d 3} e l + {^2"2 + (^ 1 ~~ ^3)"l«3}e2
+ {h^3 + (1 2 ~ ^l) w 2 w l}e 3
10.16. Derive Euler's equations of motion (22), page 256.
By the principle of angular momentum, we have
A = Tt\. W
where A is the total external torque. Writing
A = A^ + A 2 e 2 + A 3 e 3 (2)
where A 1? A 2 , A 3 are the components of the external torque along the principal axes and making
use of (1) and Problem 10.15, we find
/jwj + (I 3 — / 2 )u 2 w 3 = A t
7 2 w 2 + (h ~ h)^3 u i — A 2
7 3 w 3 + (/ 2 — /i)wi<o 2 = A 3
(3)
CHAP. 101
SPACE MOTION OF RIGID BODIES
265
FORCE FREE MOTION OF A RIGID BODY. ROTATION OF THE EARTH
10.17. A rigid body which is symmetric about an axis has one point fixed on this axis.
Discuss the rotational morion of the body, assuming that there are no forces acting
other than the reaction force at the fixed point.
Choose the axis of symmetry coincident with one of the principal axes, say the one having
direction e 3 . Then I t = 7 2 and Euler's equations become
A"i + (^3 "~ 7i)o> 2 o> 3 —
/i<o 2 + (Ii — I 3 )w 3 ui =
/ 3 «3 =
From (3), o> 3 = constant  A so that (1) and (2) become after dividing by I lt
/ 7 S  /j \
«i + f — j )Aw 2 =
h ~ h
0) 2 +
Aw, =
Differentiating (5) with respect to t and using (U), we find
•73/^2
+
h
o> 2 ~^~ k2 <°2
A 2 w 2 =
where
73^
h
Solving (7), we find o> 2 = B cos xt + C sin Kt
If we choose the time scale so that w 2 = when t = 0, then
w 2 = C sin Kt
Then from (5) we have u t = C cos Kt
Thus the angular velocity is
a = wjej + w 2 e 2 + 10363
= C cos Kt e x + C sin Kt e 2 + Ae s (11)
From this it follows that the angular velocity is
constant in magnitude equal to w = « = VC 2 + A 2
and precesses around the "3" axis with frequency
\h 7/
(1)
(2)
U)
(5)
(6)
(7)
(8)
(9)
(10)
f 2,7
2tt7!
^A
(12)
as indicated in Fig. 1013.
Note that the vector <■> describes a cone about the
"3" axis. However, this motion is relative to the
body principal axes which are in turn rotating in
space with angular velocity u.
Fig. 1013
10.18. Calculate the precession frequency of Problem 10.17 in the case of the earth rotating
about its axis.
Since the earth rotates about its axis once in a day, we have <o 3 = A = 2w radians/day. Then
the precessional frequency is from Problems 10.12 and 10.17,
/ = ~( 3 7 1 A = 77(lM = IT (.00328)(2*) = .00328 radians/day
cir \ l 1 J &tt \ Ci/ Zir
The period of precession is thus P = 1// = 305 days. The actual observed period is about 430 days
and is explained as due to the nonrigidity of the earth.
266
SPACE MOTION OF RIGID BODIES
[CHAP. 10
THE INVARIABLE LINE AND PLANE.
POLHODE, HERPOLHODE, SPACE AND BODY CONES
10.19. Describe the rotation of the earth about its axis in terms of the space and body cones.
From Problem 10.17 the angular velocity o> and angular momentum CI are given respectively by
o) = w 1 e 1 + w 2 e 2 ~^~ w 3 e 3 = C(cos Kt ej + sin Kt e 2 ) + Ae 3
CI = 7 1 « 1 e 1 + 7 1 w 2 e2 + ^3 w 3 e 3 = hC ( cos Kt e i + sm Kt e 2) + 7 3 Ae 3
Let a be the angle between o> 3 = w 3 e 3 = Ae 3 and CI. Then
and
« 3 0 = « 3 0cosa = A V/f C 2 + ZfA 2 cos a = 7 3 A2
7 3 A
V/fC 2 + 7 2 A 2
Similarly, let /? be the angle between » 3 and m. Then
» 3 • a> = «* 3  <a cos/3 = Ay/C 2 + A 2 cos /3
and cos /?
(1)
A 2
From (1) and (2) we see that
7 X C
Thus
V/fC 2 + 7fA 2
tan « =
tan a
V^ + A2
sin/8 =
VC 2 + A2
I S A
tan /3
Q
tan ft = —
A
h
h
<*)
(5)
Now for the earth or any oblate spheroid [flattened at the poles] we have 7 X < 7 3 . It follows
that a < (3.
The situation can be described geometrically by Fig. 1014. The cone with axis in the direction
CI is fixed in space and is called the space cone. The cone with axis <* 3 = u 3 e 3 is considered as
fixed in the earth and is called the body cone. The body cone rolls on the space cone so that the
element in common is the angular velocity vector <■>. Now
<l) 3 X <!>
ei e 2 e 3
A
I^C cos Kt I t C sin Kt I 3 A
= — A7 X C sin Kt e 1 + A7 X C cos Kt e 2
Thus
CI • (o> 3 X »)
(I^C cos Kt ej + 7jC sin Kt e 2 + 7 3 Ae 3 )
• (— A7 t C sin Kt ex + AI X C cos/ct e 2 )
It follows from Problem 1.21(6), page 16, that CI, o> 3 and o>
lie in one plane.
An observer fixed in space would see the vector <o
tracing out the space cone [see Fig. 1014]. An observer
fixed in the body would see the vector « tracing out the body
cone.
In the case of the earth the space cone is inside the
body cone due to the fact that l x < 7 3 . For the case of a
prolate spheroid the reverse situation is true, i.e. 7 t > 7 3 and
the space cone is outside the body cone [see Problem 10.121].
Cone
Fig. 1014
CHAP. 10]
SPACE MOTION OF RIGID BODIES
267
THE EULER ANGLES
10.20. Show by using three separate figures how the xyz coordinate system of Fig. 104,
page 257, is rotated into the x'y'z' coordinate system by successive rotations through
the Euler angles <j>, 6 and ^.
Refer to Figures 1015, 1016 and 1017. Fig.
1015 indicates a rotation through angle <f> of the x and y
axes into an X and Y axis respectively while keeping
the same z or Z axis.
In Fig. 1016 a rotation about the X axis through
angle e is indicated so that the Y and Z axes of
Fig. 1015 are carried into the Y' and Z' axes of
Fig. 1016 respectively.
In Fig. 1017 a rotation about the Z' or z' axis
through angle ^ is indicated so that the X' and Y' axes
are carried into the x' and y' axes respectively.
In the figures we have indicated unit vectors on the
x, y, z axes; X, Y, Z axes; X' , Y' , Z' axes and x', y', z'
axes by i,j,k; I, J, K; I', J', K' and i', j',k' respectively.
z or Z
Fig. 1015
Z' or z'
XorX'
Fig. 1016
Fig. 1017
10.21. Find the relationships between the unit vectors (a) i,j, k and I, J, K of Fig. 1015,
(b) I, J, K and F, J', K' of Fig. 1016, (c) I', J', K' and i', j', k' of Fig. 1017.
(a) From Fig. 1015,
i = (iI)I + (iJ)J + (iK)K
j = (j • 1)1 + (jJ)J + (jK)K
k = (kI)I + (kJ)J + (kK)K
(6) From Fig. 1016,
I = (II')I' + (IJ')J' + (I'K')K'
J = (J'I')r + (JJ')J' + (JK')K'
K = (K'I')I' + (K'J')J' + (K«K')K'
(c) From Fig. 1017,
F = (I'i')i' + (I'j')J' + (I'k')k'
J' = (J'i')i' + (J'«j')j' + (J'k')k'
K' = (K'«i')i' + (K'»j')j' + (K'k')k'
cos I — sin <f> J
sin <f> I + cos J
K
= I'
cos e J' — sin e K'
sin e J' + cos e K'
cos \p i — sin \p j
sin ^ i' + cos ^ j'
k'
268 SPACE MOTION OF RIGID BODIES [CHAP. 10
10.22. Express the unit vectors i, j,k in terms of i', j', k\
From Problem 10.21,
i = cos I — sin <p 3, j = sin I + cos <p 3, k = K
I — V, J = cos e 3' — sin K', K  sin 3' + cos K'
I' = cos0i' — sin0j', J' = sin i' + cos0j', K' — k'
Then i — cos I — sin 3 = cos <f> V — sin <p cos J' + sin <p sin K'
= cos cos i' — cos sin j'
— sin cos sin i' — sin cos cos j' + sin sin k'
= (cos cos — sin cos sin 0)i'
+ (— cos sin — sin cos cos 0)j' + sin sin k'
j = s in I + cos J = sin I' + cos cos 3' — cos sin K'
— sin cos i' — sin sin j'
+ cos cos (9 sin i' + cos cos cos 3' — cos sin * k'
— (sin cos + cos cos sin 0)i'
+ (— sin sin + cos cos cos 0)3' — cos sin k'
k — sin J' + cos JK' = sin sin i' + sin cos 3' + cos k'
10.23. Derive equations (25), page 258.
o> = co^k + Wfl I' + c^K' = 0k + Si' + 0K'
= sin (9 sin i' + sin cos 3' + cos k'
+ cos i' — sin 3' + 0k'
= (0 sin (9 sin + cos 0)i'
+ (0 sin cos — sin 0)3' + (0 cos + 0)k'
Then since o> = u x \' + w a 'j' + co 2 'k',
« x ' = w t = sin sin + cos
<V — "2 — sm * cos ~~ sin
<o s ' = <o 3 = cos +
10.24. (a) Write the kinetic energy of rotation of a rigid body with respect to the principal
axes in terms of the Euler angles, (b) What does the result in (a) become if h = hi
(a) Using Problem 10.23, the required kinetic energy is seen to be
t = £(/ lW ? + 1 2 4 + 1 A)
= ^/j(0 sin sin + cos 0) 2
+ 1/2(0 sin cos  sin 0) 2 + i/ 3 (0 cos + 0) 2
(6) If /j = 7 2 , the result can be written
T = i/ x (02 sin 2 + 02) + 1/3(0 cos + 0) 2
MOTION OF SPINNING TOPS AND GYROSCOPES
10.25. Set up equations for the motion of a spinning top having fixed point O [see Fig. 1018].
CHAP. 10]
SPACE MOTION OF RIGID BODIES
269
Let xyz represent an inertial or fixed set
of axes having origin O. Let x'y'z' represent
principal axes of the top having the same
origin. Choose the orientation of the x'y'
plane so that Oz, Oz' and Oy' are coplanar.
Then the x' axis is in the xy plane. The
line ON in the x'y' plane making an angle ^
with the x' axis is assumed to be attached
to the top.
The angular velocity corresponding to
the rotation of the x'y'z' axes with respect to
the xyz axes is
w — wie^ + w 2 e 2 ~^~ w 3 e 3
(1)
In obtaining the angular momentum we must
use the fact that in addition to the com
ponent w 3 due to rotation of the x'y'z' system
there is also the component s = se 3 = ^e 3
since the top is spinning about the z' axis.
Then the angular momentum is
Fig. 1018
O — /xwje! + 7 2 <o 2 e 2 + h^z + s ) e 3
(*)
Now if we let subscripts / and b denote the fixed system and body system respectively, we have
by Problem 6.1, page 147,
dCl
dt
<ZO I
/ dt \b
(3)
Using (1) and (2) in (3), we find
~~fo L = ih^i + (h ~ h)^2^3 + h u 2 s )^i
+ ih^2 + (^1 — ^3) w l u 3 ~ ^3 w l s ) e 2
+ {^("3 + *) + (h — ^l)"l"2} e 3
The total torque about O is
A = (fc g ) X (mg) = (le s ) x (mgk)
Since
the torque is
Then using A =
k = (k • e 1 )e 1 + (k • e 2 )e 2 + (k • e 3 )e 3 = cos (ir/2 — fl)e 2 + cos 6 e 3
= sin e 2 + cos e 3
A = —mgl(e 3 X k)
dCl
dt If
mgl sin e e x
with I t — I 2 , we find from (4) and (6),
^l w l + (^3 — A) W 2 W 3 + ^3"2 S = m 9l sm *
^1«2 + (^1 — ^3)wi"3 — ^3<0lS =
7 3 (<u 3 + s) =
U)
(5)
(6)
(7)
10.26. Express equations (7) of Problem 10.25 in terms of the Euler angles 6 and <£ of
Fig. 1018.
The components o> v « 2 , w 3 can be obtained from Problem 10.23 by formally letting ^ = 0.
We find
Ul — 0, u 2 = sm 9, a5 3 = COS 6
Then equations (7) of Problem 10.25 become
(1)
270 SPACE MOTION OF RIGID BODIES [CHAP. 10
h '°' + (^3 — h)$ 2 sm e cos e + h$ s sm 6 = m ^ sin e
hC<P sin<9 + $e cose) + (7j — I 3 )e^>cose — I 3 es =0 V (2)
7 3 ( cos o — $e sin e + s) —
The quantities <£, and s are often known as the magnitudes of the angular velocity of precession,
of nutation and of spin respectively.
10.27. Prove that the equations (2) of Problem 10.26 can be written as
(a) he  7i<J) 2 sin cos + Z" 3 A</>sin0 = mgl sin e
(b) h{'4 sin e + 2j>6 cose)  hA8 =
where A is a constant.
From the third equation in (7) of Problem 10.25,
w 3 + s = A or s = A — « 3 (1)
Then substitution into the first and second of equations (7) yields
7i"i ~ 7 1 w 2 w3 + 7 3 « 2 A — m 9l sin e W
/ 1 6j 2 + 7 1 <o 1 « 3 — /3W1A = W
Using the results (I) of Problem 10.26, we find that equations (2) and (3) reduce to the required
equations.
10.28. (a) Find the condition for steady precession of a top.
(b) Show that two precessional frequencies are possible.
Since is constant so that V = 0, we have from Problem 10.27(a),
(Irf 2 cos e — 7 3 A<£ + mgl) sin =
or Irf 2 cos — I 3 A$ + mgl =
7 3 A ± V7fA 2  Amglh cos e
from which <t> = g^^ I >
Thus there are two frequencies provided that
/fA 2 > Amglli cos e (2)
If 7 3 A 2 = 4mgll x cos only one frequency is possible.
If A is very large, e.g. if the spin of the top is very great, then there are two frequencies,
one large and one small, given by
I 3 A/(I lC ose), mgl/I 3 A W
10.29. Prove that
(a) %h(e 2 + <J» 2 sin 2 8) + / 3 A 2 + mgl cos e  constant = E
(b) h4> sin 2 d + hA cos e = constant = K
and give a physical interpretation of each result.
(a) Multiply equations (7) of Problem 10.25 by Wl , co 2 and <* 3 + s respectively, and add to obtain
/ifttji! + *>2«2) + 7 3("3 + *)(«S +«) = m ^ Sin ' *
which can be written as
f t {#M + 4) + W a * + s > 2 > = f t (~ m 9lcose)
Integrating and using <* 3 + s = A as well as the results Wl = * and <o 2 = sine, we find
ll^e* + <£ 2 sin 2 0) + 7 3 A 2 + mgl cos e = E CO
CHAP. 10] SPACE MOTION OF RIGID BODIES 271
where E is constant. The result is equivalent to the principle of conservation of energy,
since the kinetic energy is
T = AJ^fl 2 + ^ 2 sin2 e) + %I 3 A* (2)
while the potential energy is V = mgl cos e (8)
and T + V — E is the total energy.
(6) Multiplying the result of Problem 10.27(6) by sin e,
I^sirfo + 11 x <t>6 sin e cos e — I 3 Ae sine =
which can be written
jt(/i0 sin 2 + I 3 A costf) =
Integrating, Irf sin 2 + I 3 A cos 6 — constant = K (4)
To interpret this result physically, we note that the vertical component of the angular
momentum is Irf sin 2 e + I 3 A cos e [see Problem 10.123], and this must be constant since the
torque due to the weight of the top has zero component in the vertical direction.
10.30. Let u = cos 6. Prove that:
(a) u 2 = {a/3u)(lu 2 ) {ySu) 2 = f(u)
where a = 2(E  ihA 2 )/h, p = 2mgl/h, y = K/I u 8 = hA/h;
(6) t = j
du
7m
+ constant
(a) From Problem 10.29,
i/ 1 (ff 2 + <p 2 sin 2 e) + ±I 3 A* + mgl cos 6 = E (1)
Irf sin 2 6 + I 3 A cos = K {2)
K — I 3 A cos e
From (2), * = / lSin 2, <*>
Substituting this into (I),
(K  I 3 A cos 0)2
i' 1 ' 2 + 2 / lS in2 g + ¥* A + mgU ° se = E
Letting u = cos * so that u = —sin 8 h and sin 2 9 = 1 — u 2 , this becomes
y.r^ + 27.(1^) + "*"" = £ * 7 ^ 2
Thus * + (^p) 2 + 2rog '^" 2) = «&^a v _ 1 / 1 A,
which can be written as
w 2 = (a  fiu)(l  u*)  (y  8u) 2 = /(«) (4)
where a = (2tf  I^A 2 )/^, p = 2mgl/I 1 , y = K/I lt 8 = /aA/^ (5)
Note that with this notation {3) can be written
. _ y  8u . .
* ~ T^ 2 " (<?)
272
SPACE MOTION OF RIGID BODIES
[CHAP. 10
(b) From the result of (a) we have, since u > 0,
Integrating,
£ =
s
or
+ c
<&*
(7)
The integral can be evaluated in terms of elliptic functions which are periodic.
10.31. (a) Prove that 6 = at those values of m for which
f(u) = («/8M)(lM 2 )(y8M) 2 
(b) Prove that the equation in (a) has three real roots u u u 2 ,u 3 but that in general
not all the angles corresponding to these are real.
(a) From Problem 10.30(a), u 2 = f(u) = (a  £m)(1 ~ « 2 )  (y  8u) 2
Since u = sin e 0, it follows that e = where m = or /(«) = 0.
roots of the equation
/( M ) = ( a  fiu)(l  m 2 )  (y  S«)2 =
(b) Equation (1) can be written as
f(u) = /3tt 3  (S 2 + a)u 2 + (2yS  /3)u + a  y 2
(1)
Thus 6 = at the
(2)
Since /? > 0, it follows that
/(I)
= (yS) 2 , /(l)
Thus there is a change of sign from
— to + as u goes from 1 to °°, and
consequently there must be a root, say
u 3 , between 1 and » as indicated in
Fig. 1019.
Now we know that in order for
the motion of the top to take place we
must have f(u) = u 2 s o. Also, since
^ e ^ ir/2, we must have = tt = 1.
It thus follows that there must be two
roots m x and u 2 between and 1, as
indicated in the figure.
It follows that in general there are
two corresponding angles e x and e 2 such
that cos 0i
cos 2 — u 2 ^ n special
cases it could happen that u x
U<l = Ug = 1.
Fig. 1019
10.32. Give a physical interpretation of the results found in Problem 10.31.
The fact that there are two roots % and u 2 corresponding to 6 X and 2 respectively, shows
that the motion of the top is such that its axis always makes an angle e with the vertical which
lies between e y and e 2 . This motion, which is a bobbing up and down of the axis between the
limits <?! and 2 , is called nutation and takes place at the same time as the precessional motion
of the axis of the top about the vertical and the spinning of the top about its axis. Because
the motion can be expressed in terms of elliptic functions [see Problem 10.104], we can show
that it is periodic.
In general the tip of the axis of the top will describe one of various types of curves such as
indicated in Figs. 1020, 1021 and 1022. The type of curve will depend on the root of the
equation [see equation (6) of Problem 10.30]
<t> =
_ y — Su
 u 2
(1)
CHAP. 10]
SPACE MOTION OF RIGID BODIES
273
If this root given by y/8 is greater than u 2 , the curve of Fig. 1020 occurs. If it is the same as
u 2 , the curve of Fig. 1021 obtains. If it is between u x and u 2 the curve of Fig. 1022 occurs. Other
cases can arise if the root is the same as u x or is less than u x [see Problems 10.124].
y/8 > u 2
Fig. 1020
y/8 = u 2
Fig. 1021
ul < y/8 < u 2
Fig. 1022
Aside from the general motion which is made up of nutation and precession, there are
various special cases which can arise. One of these is the case of steady precession with no
nutation [see Problem 10.28]. In this case u x = u 2 so that e x ~ e 2 or e = constant. Another
case is the "sleeping top" which occurs where u 2 = u 3 = 1 and the axis of the spinning top always
remains vertical [see Problem 10.36].
MISCELLANEOUS PROBLEMS
10.33. If T is the total kinetic energy of rotation of a rigid body with one point fixed,
prove that dT/dt  <o • A where all quantities refer to the body principal axes.
Multiplying both sides of the Euler equations (3) of Problem 10.16 by w 1 ,w 2 , <o 3 respectively
and adding, we obtain
^l w l w l ~^~ •^2 (0 2 w 2 ~f~ ^3 W 3 W 3 = "l^l ~^~ c0 2^2 ~^~ w 3^3 (7)
But
and
Thus (1) becomes
t • ^ t • lt • _l^/r2j_r2.r 2 \— dT
iiCOjWj + i 2 O) 2 0) 2 + '3 W 3 W 3 — o di lWl 2< ° 2 3< ° 3 ' — ~dt
0)^ + 0) 2 A 2 + "3^3 = ("l^l + «2 e 2 + <">3 e 3) * ( A l e l + A 2 e 2 + A 3 e 3 )
= 6> • A
dT/dt
at • A
(2)
(3)
10.34. (a) Prove that if there are no forces acting on a rigid body with one point fixed,
then the total kinetic energy of rotation is constant, (b) Thus prove that © • O =
IT — constant.
(a) Since there are no forces, A = 0. Then by Problem 10.33, dT/dt  or T = constant.
(6) Since O = /jo^ej + I 2 u 2 e 2 + 730)363 and «• = o^e^ + o> 2 e 2 + 0)363,
«* • O = / x o)f + 7 2 w + / 3 o>3 = 2T — constant
10.35. Find the precession frequency of Problem 10.17 in terms of the kinetic energy and
angular momentum of the rigid body.
The kinetic energy is
T = i(/,o) 2 + 7 2 o) + I 8 «) = 4(/ lW ? + h4 + 7 3W ) = ¥h° 2 + J s A2 )
so that
/jC 2 + 7 3 A2 = 2T
(1)
274
SPACE MOTION OF RIGID BODIES
[CHAP. 10
The angular momentum is
O = Il03 1 e 1 + /2 w 2 e 2 + ^3 w 3 e 3 = ^l w l e l + ^l"2 e 2 + ^3 w 3 e 3
= 7 1 C(cos Kt e l + sin id e 2 ) + ^3Ae 3
so that = Q = V/fC 2 + /A2 or
I 2 C 2 + 1%A 2 = S22
Solving (i) and (2) simultaneously, we find
277 3  fi2 fi 2 _ 277,
(72  A 2 
hdaIt)' A h(hh)
Then from Problem 10.17, equation (12), the precession frequency is
2^/i \
\(n*2Ti 1 )(i 3 i l )\
(2)
(3)
04)
10.36. Find the condition for a "sleeping top".
For a "sleeping top" we must have e = and 5 = 0, since the axis must remain vertical
and no nutation can take place. Then from Problem 10.29,
hA
K,
I 3 A 2  2(Emgl)
Also, from Problem 10.30 we have a = 2mgl/I l , = 2mgl/I v y = l 3 All x , 8 = I 3 A/I V Thus a  ($
and y = 8, and so
/(h) = (a  j8u)(l  h2)  (y  8h)2 = «(1  u)(l  it*)  y 2 (l  u)2 = (1  h)2 [«(1 + u)  y2]
It follows that /(h) = has a double root at u — 1, while the third root is given by
v 2
1 =
JA«
 1
2mgll 1
Then the top will "sleep" if this root is greater than or at most equal to 1, so that
A2 ^ Amglljll
Of course, even though this condition may apply at the beginning, energy will in practice
be diminished due to friction at the support so that after some time we will have A 2 < Amglljl^.
In such case precession combined with nutation will be introduced. Further loss of energy will
ultimately cause the top to fall down.
10.37. Find the torque needed to rotate a rectangular plate of sides a and b [see Fig. 1023]
about a diagonal with constant angular velocity ©.
By Problem 10.9 the principal moments of inertia of the
plate at the center O are given by
l x = ^M<fi t I 2 = ^Mb*, I z = ^M(a 2 + b*)
We have
a> = (wi)i 4 (»*j)j
Thus
Substituting (1) and
Va 2 + 6 2
Va 2 + 6 2
(1)
(2)
— osb
w 2
V» 2 + 6 2 '
Va 2 + 6 2 '
into Euler's equations
/jWi + (7 3 — 72) W 2 W 3 = A l
I 2 a 2 + (/ x — 73)<°3 W 1 = A 2
/ 3 <J) 3 + (J 2 — Ji)"i"2 = A 3
Fig. 1023
w 3
CHAP. 10]
SPACE MOTION OF RIGID BODIES
275
M(b 2 — a z )abu 2 „,, , . , A , ^ ^ .
we find Aj = 0, A 2 = 0, A 3 = U( 2 + b*) ■ Thus the re( l uired torque about O is
M(&2  cfi)ab<J
A ~ 12(a2 + 62)
Note that if the rectangular plate is a square, i.e. if a — b, then A = 0.
(4)
Supplementary Problems
GENERAL MOTION OF RIGID BODIES IN SPACE
10.38. Find the number of degrees of freedom of (a) a sphere free to roll on a plane, (b) an ellipsoid
free to rotate about a fixed point, (c) an airplane moving in space.
Ans. (a) 3, (b) 3, (c) 6
10.39. In Fig. 1024 a displacement of a tetrahedron in
space is indicated. Show directly that the dis
placement can be accomplished by a translation
plus a rotation about a suitable axis, thus
illustrating Chasle's theorem [page 224] for
space.
10.40. Give an illustration similar to that of Problem
10.39 involving a rigid body whose surfaces are
not plane surfaces.
10.41. Derive the result of Problem 10.2, page 259,
without using Problem 6.1, page 147.
Fig. 1024
ANGULAR MOMENTUM. KINETIC ENERGY. MOMENTS AND PRODUCTS OF INERTIA
10.42. A rigid body consists of 3 particles of masses 2,1,4 located at (1,1,1), (2,0,2), (—1,1,0) respec
tively. Find the angular momentum of the body if it is rotated about the origin with angular
velocity » = 3i  2j + 4k. Arts. 6j + 42k
10.43. Determine the (a) moments of inertia about the x, y and z axes and (b) the products of inertia
for the rigid body of Problem 10.42.
Ans. (a) I xx = 12, I yy = 16, I„ = 16; (6) I xy = 6, I yz = 2, 1„ = 6
10.44. What is the kinetic energy of rotation for the system of Problem 10.42?
Ans. 180
10.45. Find the (a) moments of inertia and (6) products of inertia of a uniform rectangular plate
ABCD of sides AB = a and AD — b taken about axes AB, AD and the line perpendicular to
the plate at B.
Ans. (a) I xx = iM62, Iyy = l M a?, I gz = ±M(a? + 62)
(&) hy = ~lMab, l yz = 0, I xz =
calling axes through AB and AD the x and y axes respectively.
10.46. Find the (a) moments of inertia and (6) products of inertia of a cube of side a taken about x,y, z
axes coinciding with three intersecting edges of the cube.
Ans. (a) I xx = I yw = I zz  %Ma2, (b) I xy = I yz = I xz  \Ma*
10.47. Find the (a) angular momentum and (6) kinetic energy of rotation of the cube of Problem 10.46
about the point of intersection O of the three edges if the cube has an angular velocity
a, = 2i + 5j  3k about O. Ans. (a) ^MaHlOi + 43j  45k), (6) 185MaV12
276 SPACE MOTION OF RIGID BODIES [CHAP. 10
10.48. Find the (a) moments of inertia and (b) products of inertia of the uniform solid sphere
x 2 + y 2 + z 2 = a 2 in the first octant, i.e. in the region *g0, 2/ = 0, z ^ 0.
Ans. (a) I xx = I m = I zz = §Ma 2 , (b) I xy = I yz = I xz = 2MaV5ir
PRINCIPAL MOMENTS OF INERTIA. PRINCIPAL AXES. ELLIPSOID OF INERTIA
10.49. Prove that the principal moments of inertia for a system consisting of two particles of masses
m l and m 2 connected by a massless rigid rod of length I are I t = J 2 — m l m 2 ^/{m l + m 2 ), I z = 0.
10.50. Find the (a) principal moments of inertia and (b) directions of the principal axes for the system
of Problem 10.42.
Ans. (a) I x = 18, I 2 = 13  t/T3 , I 3 = 13 + \/73
(b) j + k, £(/ + V73)iJ + k, j(i_/73)i_j + k>
10.51. Determine the (a) principal moments of inertia and
(b) directions of the principal axes for right triangle ABC
of Fig. 1025 about point C.
10.52. Find the principal moments of inertia at the center of a
parallelogram of sides a and b and acute angle a.
10.53. Find the (a) principal moments of inertia and (6) direc
tions of the principal axes for the cube of Problem 10.46.
Ans. (a) Zj = 7 2 = flilfa 2 , I 3 = ±Ma 2
(6) Axis associated with I s is in the direction of the diagonal from the origin. Axes
associated with J x and I 2 have any mutually perpendicular directions in a plane perpen
dicular to this diagonal.
10.54. Find the principal moments of inertia of a uniform cylinder of radius a and height h.
Ans. J a = I 2 = jLM(3a 2 + h 2 ), I 3 = \Ma 2
10.55. Obtain the principal moments of inertia and directions of principal axes for a rectangle of
sides a and b by using Problem 10.45 and equations (11), page 255. Compare with Problem 10.9,
page 262.
10.56. Find the lengths of the axes of the ellipsoid of inertia corresponding to the rectangle of
Problem 10.55. Ans. 4^3/Ma 2 , Ay/Z/Mb 2 , 4^S/M(a 2 + b 2 )
10.57. Find the lengths of the axes of the ellipsoid of inertia corresponding to the cube of Problem 10.46.
Ans. 4\/3/llMa 2 , 4V3/llMa 2 , 2 v / 6/Ma 2
10.58. Prove that the ellipsoid of inertia for a regular tetrahedron is a sphere and determine its radius.
10.59. If I V I 2 ,I 3 are the principal moments of inertia, prove that
I, ^ I 2 + h, h ^ h + h. h ^ h + h
10.60. Under what conditions do any or all of the equality signs hold in Problem 10.59?
10.61. Prove that if a rigid body is a solid of revolution about a line L, then L is a principal axis
corresponding to any part of L.
10.62. Suppose that a rigid body is symmetrical about a plane P. Prove that if L is a line perpendicular
to P at point O, then L is a principal axis corresponding to point O.
EULER'S EQUATIONS OF MOTION
10.63. A rigid body having one point O fixed and no external torque about O, has two equal principal
axes of inertia. Prove that it must rotate with constant angular velocity.
CHAP. 10] SPACE MOTION OF RIGID BODIES 277
10.64. Write Euler's equations for the case of plane motion of a rigid body and discuss their physical
significance.
10.65. Solve the problem of a compound pendulum by using Euler's equations.
10.66. Describe how Euler's equations can be used to discuss the motion of a solid cylinder rolling down
an inclined plane.
10.67. Write Euler's equations in case the axes are not principal axes.
FORCE FREE MOTION. INVARIABLE LINE AND PLANE.
POLHODE, HERPOLHODE, SPACE AND BODY CONES
10.68. If two principal moments of inertia corresponding to the fixed point about which a rigid body
rotates are equal, prove that (a) Poinsot's ellipsoid is an ellipsoid of revolution, (6) the polhode
is a circle and (c) the herpolhode is a circle.
10.69. Discuss the (a) invariable line and plane, (6) polhode and herpolhode and (c) space and body
cones for the case of a rigid body which moves parallel to a given plane, i.e. plane motion of a
rigid body.
10.70. (a) How would you define the instantaneous axis of rotation for space motion of a rigid body?
(6) What is the relationship between the instantaneous axis of rotation and the space and body
cones?
10.71. Prove that relative to its center of mass the axis about which the earth spins in a day will
rotate about an axis inclined at 23.5° with respect to it in 25,780 years.
THE EULER ANGLES
10.72. Using the notation of Problem 10.20, page 267, find: (a) I, J,K in terms of i,j,k; (6) I\J',K' in
terms of I,J,K; (c) i',j',k' in terms of I',J',K'.
Ans. (a) I = cos <f> i + sin <f> j, J = —sin $ i + cos <f> j, K = k
(6) I' = I, J' = cos e J + sin e K, K' = —sin e J + cos e K
(c) i' = cos ^ F + sin ^ J', J' = —sin f I' + cos $ J', k' = K'
10.73. Prove the results u x = 9 cos <f> + ^ sin e sin <j>
a y = sin <f> — ^ sin e cos
u g = $ + yp cos *•
10.74. If I x = I 2 = I3, prove that the kinetic energy of rotation of a rigid body referred to principal
axes is T = \I X (4? + o 2 + j< 2 + 2w cos e).
MOTION OF SPINNING TOPS AND GYROSCOPES
10.75. A top having radius of gyration about its axis equal to 6 cm is spun about its axis. The spinning
point is fixed and the center of gravity is on the axis at a distance 3 cm from this fixed point.
If it is observed that the top precesses about the vertical at 20 revolutions per minute, find the
angular speed of the top about its axis. Ans. 3.10 rev/sec or 19.5 rad/sec
10.76. A uniform solid right circular cone of radius a and height h is spun so that its vertex is fixed
and its axis is inclined at a constant angle a with the vertical. If the axis precesses about the
vertical with period P, determine the angular speed of the cone about its axis.
10.77. Work Problem 10.76 if the cone is surmounted by a uniform solid hemisphere of radius a and the
same density.
10.78. Explain physically why the spin axis of the gyroscope of Figures 106 and 107, page 258,
should maintain its direction.
278
SPACE MOTION OF RIGID BODIES
[CHAP. 10
10.79. Explain how a gyroscope can be used to enable a ship, airplane, submarine or missile to follow
some specified course of motion.
MISCELLANEOUS PROBLEMS
10.80. A uniform solid cube of side a and mass M has its edges lying on the positive x, y and z axes
of a coordinate system with vertex at the origin O. If it rotates about the z axis with constant
angular velocity w, find the angular momentum. Arts, — J^Ma 2 <o(3i + 3j — 8k)
10.81. Find the moment of inertia of a uniform solid cone of radius a, height h and mass M about
(a) the base, (b) the vertex. Arts, (a) ^Ma 2 , (b) ^M{2h 2 + a 2 )
10.82. Find the principal moments of inertia at the center of a uniform elliptical plate having semimajor
axis a and semiminor axis b. Ans. I t = %Mb 2 , I 2 = \Ma 2 , I 3 = \M{a 2 + b 2 )
10.83. A top has the form of a solid circular disk of radius o and mass M
with a thin rod of mass m and length I attached to its center
[see Fig. 1026]. Find the angular velocity with which the top
should be spun so as to "sleep". Assume that the base point O
is fixed.
10.84. Work Problem 10.83 for a cone of radius a, height h and mass M.
10.85. Work Problem 10.83 for a cone of radius a, height h and mass M
surmounted by a hemisphere of radius a and mass m.
10.86. A coin of radius a is set spinning about a vertical axis with
angular velocity o> [see Fig. 1027]. Prove that the motion is stable
if to 2 > Agla.
10.87. Suppose that the coin of Problem 10.86 is spun with angular speed
s about a diameter which is inclined at an angle a with the vertical
and which is fixed at point O. Assuming there is no nutation,
find the angular speed with which the coin precesses about the
vertical.
10.88. Discuss how gyroscopes can be used to control the motions of a ship
on a stormy sea.
Fig. 1026
Fig. 1027
10.89. The vertex of a uniform solid cone of radius a, height h and mass M is fixed at point O of a
horizontal plane. Prove that if the cone rolls on the plane with angular velocity <o about an axis
, . . ..... 3Mh 2 (a? + 6fc 2 )co 2
perpendicular to the plane through O, then the kinetic energy of rotation is 40(a 2 + h 2 ) '
10.90. Explain how the principal axes of a rigid body can be found if the direction of one of the
principal axes is known.
10.91. A uniform solid cone has the radius of its base equal to twice
its altitude. Prove that the ellipsoid of inertia corresponding
to its vertex is a sphere.
10.92. Explain how a gyroscope can be used as a compass, often
called a gyrocompass.
10.93. A dumbbell consists of two equal masses M attached to a
rod ABC of length I and negligible mass [see Fig. 1028].
The system rotates about a vertical axis DCE with constant
angular velocity «* such that the rod makes a constant
angle 8 with the vertical. Prove that the angular momentum
Q of the system describes a cone of angle irl2 — e about u
and has magnitude \Ml 2 u sin e .
CHAP. 10]
SPACE MOTION OF RIGID BODIES
279
10.94. (a) Prove that the magnitude of the torque needed to keep the system of Problem 10.93 in motion
is \Ml 2 v> 2 sin2<?  (&) What is the direction of the torque?
10.95. Work (a) Problem 10.93 and (6) Problem 10.94 if the rod ACB has mass m.
10.96. A thin solid uniform circular plate of radius a has its center
attached to the top of a thin fixed vertical rod OA [see
Fig. 1029]. It is spun with constant angular speed w about
an axis which is inclined at angle a with the normal OB to
the plate, (a) Prove that the angular velocity vector «*
precesses about the normal OB with period 2tt/(u cos a).
(b) Prove that the axis OB describes a space cone with period
2tt/(<o \/1 + 3 cos 2 a) .
10.97. In Problem 10.96 find the angle through which the plate
turns during the time it takes OB to describe the space cone.
10.98. Find the principal moments of inertia of a uniform solid cone
of radius a, height h and mass M taken about the (a) vertex,
(6) center of mass.
Ans. (a) I x = h = ^M(a 2 + Ah 2 ), I 3 = ^Ma 2
(b) h=h= i>M(h* + 4a 2 ), 7 3 = J^Ma 2
Fig. 1029
10.99. A compound pendulum of mass M oscillates about a horizontal axis which makes angles a, /?, y
with respect to the principal axes of inertia. If the principal moments of inertia are 7 1 ,7 2 , 7 3
respectively and the distance from the center of mass to the axis of rotation is I, prove that for
small oscillations the period is 2v^Mgl/I where I = Ml 2 + 1^ cos 2 a + I 2 cos 2 (3 + / 3 cos 2 y.
10.100. Find the period of small oscillations of a uniform solid
cone which rotates about a horizontal axis attached to
the vertex of the cone.
10.101. An elliptical plate [see Fig. 1030] having semimajor
and semiminor axes of lengths a and 6 respectively is
rotated with constant angular speed « about an axis
making a constant angle a with the major axis. Find
the torque required to produce this motion.
10.102. Work Problem 10.101 if the elliptical plate is replaced
by an ellipsoid. Fig. 1030
10.103. Given Euler's equations of motion for a rigid body having zero external torque about a fixed
point O, i.e.,
I^X + (/ 3 — jT 2 )u 2 <o 3 = 0, 7 2 «2 + Cl ~" ^3) W 3 W 1 = 0, Z 3 « 3 + (/ 2 — ^l)«l"2 =
prove that
and
^i w i ~^~ ^2 W 2 ~t" ^3 U 3 = constant — 2T
/?«i + /u + 7 3 <o = constant = H 2
10.104. Prove from Problem 10.103 that a lt « 2 and « 3 satisfy a differential equation of the form
dy/dx = V(l — # 2 )(1 ~ k 2 x 2 ), and thus show that the angular velocity can be expressed in terms
of elliptic functions.
10.105. Find the moment of inertia of a uniform solid cone of radius a, height h and mass M about a
line which lies in its surface. Ans. ^Ma 2 (a 2 + 6h 2 )/(a 2 + h 2 )
10.106. The moments and products of inertia of a rigid body about the x, y and z axes are I xx = 3,
I yy = 10/3, I zz = 8/3, I xy = 4/3, I ^ = — 4/3, I yz = 0. Find (a) the principal moments of inertia and
(6) the directions of the principal axes.
Ans. (a) I x = 3, J 2 = 2, 7 3 = 4
(6) ei = i  2j  2k, e 2 = 2i + j  2k, e 3 = 2i  2j + k
280
SPACE MOTION OF RIGID BODIES
[CHAP. 10
10.107. A cone having semivertical angle o rolls with constant angular speed w on a horizontal plane with
its vertex fixed at a point O. Prove that the axis of the cone rotates about the vertical axis
through O with angular speed w tan a.
10.108. A horizontal plane rotates about a vertical axis with constant angular velocity <■». A uniform
solid sphere of radius a is placed on this plane. Prove that the center describes a circle with
angular velocity given in magnitude by w.
10.109. Work Problem 10.108 if the sphere is not necessarily of constant density.
Ans. aK 2 /(K 2 + a 2 ) where K is the radius of gyration about a diameter
10.110. Show how to find the relative maximum and minimum distances from the origin to the ellipsoid
* = Ax 2 + By 2 + Cz 2 + Dxy + Eyz + Fxz  1.
[Hint. Maximize or minimize the function * = x 2 + y 2 + z z subject to the condition * = 1.
To do this use the method of Lagrange multipliers, i.e. consider the function G = * + X* where
X is the (constant) Lagrange multiplier and set dG/dx, dG/dy, dG/dz equal to zero.]
10.111. Explain the relationship of Problem 10.110 to the method of page 255 for obtaining principal
moments of inertia and directions of principal axes.
10.112. (a) Find the relative maximum and minimum distances from the origin to the ellipsoid
9x 2 + 10y 2 + Sz 2 + Axy  4scz = 3.
(b) Discuss the connection of the results of (a) with those of Problem 10.106.
10.113. Find the moment of inertia of the system of particles of Problem 10.42 about a line through the
point (2, — 1, 3) in the direction 3i — 2j + 4k.
10.114. Prove that the motion of the "sleeping top" of Problem 10.36 is stable if A 2 ^ Amglljl^
10.115. Find the moment of inertia of the lemniscate r 2  a 2 cos 2e about the z axis
10.116
Ans. jfMa 2
A plane rigid body (lamina) has an xy and x'y' coordinate system with common origin O such
that the angle between the x and x' axes is a [see Fig. 1031]. Prove that
(a) I x > x > = I xx cos 2 a — 2I xy sin a cos a + I yy sin 2 a
I rT sin 2 a + 21 rv sin a cos a + /„„ sin 2 a
10.117. Use Problem 10.116 to prove that
and give a physical interpretation.
10.118. Referring to Problem 10.116, find an expression for I x  y  in terms of J xx ,I xy ,I yy and a.
10.119. Use the results of Problems 10.116 and 10.118 to prove that for a plane region having moments
and products of inertia defined by I xx , I xy> I yy corresponding to a particular xy coordinate system,
the principal axes are obtained by a rotation of these axes through an angle a given by
tan 2a = I xy /(I yy — I xx ).
10.120. Prove that the lengths of the principal axes in Problem 10.116 are given by
CHAP. 10] SPACE MOTION OF RIGID BODIES 281
10.121. Discuss Problem 10.19, page 266, if l x > / 3 .
10.122. Find the moment of inertia of a uniform semicircular wire of mass M and radius a about its
center. Ans. 2M(ir — 2)a 2 /v
10.123. Prove that the expression on the left side of equation (4) in Problem 10.29 is the vertical
component of the angular momentum.
10.124. Discuss Problem 10.32 if the root of equation (1) is (a) equal to u lf (b) less than u v
10.125. A rigid body consists of 3 particles of masses m u ra 2 and ra 3 . The distance between m l and m 2 ;
m 2 and ra 3 ; m s and m l are l 12 , hz an d Z 3 i respectively. Prove that the moment of inertia of the
system about an axis perpendicular to the plane of the particles through their center of mass is
given by .2 , ,2 , ,2
Wi»»2*i2 + m 2 m 3 l 2 3 + m^m^
mi + m 2 + m 3
10.126. Derive a "parallel axis theorem" for products of inertia and illustrate by means of an example.
10.127. Prove that the principal moments of inertia of a triangle of sides a,b,c and mass M about the
center of mass are given by
I t = I 2 = ^(«2 + fe2 + C 2 ± 2 Va 4 + 6 4 + C4  tt 2&2 _ fc2 c 2 _ C 2 a 2 ), / 3 = ff (« 2 + & 2 + C 2 )
10.128. A coin of radius 1.5 cm rolls without slipping on a horizontal table such that the plane of the
coin makes an angle of 60° with the table. If the center of the coin moves at a speed of 3 m/sec,
prove that the coin moves in a circular path and find its radius. Ans. 2.5 m
Chapter 11 LAGRANGE'S
EQUATIONS
GENERAL METHODS OF MECHANICS
Up to now we have dealt primarily with the formulation of problems in mechanics by
Newton's laws of motion. It is possible to give treatments of mechanics from rather
general viewpoints, in particular those due to Lagrange and Hamilton.
Although such treatments reduce to Newton's laws, they are characterized not only
by the relative ease with which many problems can be formulated and solved but by their
relationship in both theory and application to such advanced fields as quantum mechanics,
statistical mechanics, celestial mechanics and electrodynamics.
GENERALIZED COORDINATES
Suppose that a particle or a system of N particles moves subject to possible constraints,
as for example a particle moving along a circular wire or a rigid body moving along an
inclined plane. Then there will be a minimum number of independent coordinates needed
to specify the motion. These coordinates denoted by
q\, <?2, . . .,q n (1)
are called generalized coordinates and can be distances, angles or quantities relating to
them. The number n of generalized coordinates is the number of degrees of freedom
[see page 165].
Many sets of generalized coordinates may be possible in a given problem, but a
strategic choice can simplify the analysis considerably.
NOTATION
In the following the subscript a will range from 1 to n, the number of degrees of
freedom, while the subscript v will range from 1 to N, the number of particles in the
system.
TRANSFORMATION EQUATIONS
Let r„ = x v i + Vvj + z v k be the position vector of the vth particle with respect to an
xyz coordinate system. The relationships of the generalized coordinates (1) to the 1 position
coordinates are given by the transformation equations
X v = X v (qi, #2, • • • , Qn, t)
ijv = Vv{qi, q%, . . . , qn, t) [ (2)
z v = z v (qi, q 2 , . . ., qn, t)
where t denotes the time. In vector form, (2) can be written
r v  r v {qi, q%, . . . , q n , t) (#)
The functions in (2) or (3) are supposed to be continuous and to have continuous derivatives.
282
CHAP. Ill
LAGRANGE'S EQUATIONS 283
CLASSIFICATION OF MECHANICAL SYSTEMS
Mechanical systems can be classified according as they are scleronomic or rheonomic,
holonomic or nonholonomic, and conservative or nonconservative as defined below.
SCLERONOMIC AND RHEONOMIC SYSTEMS
In many mechanical systems of importance the time t does not enter explicitly in the
equations (2) or (3). Such systems are sometimes called scleronomic. In others, as for
example those involving moving constraints, the time t does enter explicitly. Such systems
are called rheonomic.
HOLONOMIC AND NONHOLONOMIC SYSTEMS
Let q it q 2 , ...,q n denote the generalized coordinates describing a system and let t
denote the time. If all the constraints of the system can be expressed as equations having
the form <j>{qi, q 2 , . . . , q n , t) = or their equivalent, then the system is said to be holonomic;
otherwise the system is said to be nonholonomic. Compare page 170.
CONSERVATIVE AND NONCONSERVATIVE SYSTEMS
If all forces acting on a system of particles are derivable from a potential function
[or potential energy] V, then the system is called conservative, otherwise it is noncon
servative.
KINETIC ENERGY. GENERALIZED VELOCITIES
The total kinetic energy of the system is
T = I 5>„r2 (*)
6 v=l
The kinetic energy can be written as a quadratic form in the generalized velocities q a .
If the system is scleronomic [i.e. independent of time * explicitly], then the quadratic form
has only terms of the form a^haq&. If it is rheonomic, linear terms in q a are also present.
GENERALIZED FORCES
If W is the total work done on a system of particles by forces F„ acting on the vth
particle, then B
dW = *i£$adq a (5)
<x = l
N
dr v
where *« = 2 Fv * T^~ ^
v = l
dq a
is called the generalized force associated with the generalized coordinate q a . See
Problem 11.6.
LAGRANGE'S EQUATIONS
The generalized force can be related to the kinetic energy by the equations [see
Problem 11.10]
±(9T) iT = * 8 (7)
dt\dq a J dq a
284 LAGRANGE'S EQUATIONS [CHAP. 11
If the system is conservative so that the forces are derivable from a potential or potential
energy V, we can write (7) as
d(BL\ _ BL =
dt\dq a J dq a
where L = T  V (9)
is called the Lagrangian function of the system, or simply the Lagrangian.
The equations (7) or (8) are called Lagrange's equations and are valid for holonomic
systems which may be scleronomic or rheonomic.
If some of the forces in a system are conservative so as to be derivable from a potential
V while other forces such as friction, etc., are nonconservative, we can write Lagrange's
equations as
*(*£)  *k = *; (10)
dt\dq a J dq a
where LTV and $>L are the generalized forces associated with the nonconservative
forces in the system.
GENERALIZED MOMENTA
We define at
dq a
to be the generalized momentum associated with the generalized coordinate q a . We often
call p a the momentum conjugate to q a , or the conjugate momentum.
If the system is conservative with potential energy depending only on the generalized
coordinates, then (11) can be written in terms of the Lagrangian L = T  V as
BL
Pa
Bq a
(12)
LAGRANGE'S EQUATIONS FOR NONHOLONOMIC SYSTEMS
Suppose that there are m equations of constraint having the form
VAadqa + Adt = 0, ^Badq a + Bdt = 0, ... (13)
a a
or equivalents ^A a q a + A = 0, ^B a q a + B = 0, ... (U)
a «
We must of course have m < n where n is the number of coordinates q a .
The equations (13) or (U) may or may not be integrable so as to obtain a relationship
involving the q a 's. If they are not integrable the constraints are nonholonomic or non
integrable; otherwise they are holonomic or integrable.
In either case Lagrange's equations can be replaced by
d_ (dT\ _ aT = ^ + XiAa + XzBa + ... (i 5)
dt\dq a J dq a
where the m parameters Ai, A 2 , ... are called Lagrange multipliers [see Problem 11.18].
If the forces are conservative, (15) can be written in terms of the Lagrangian
L = TV as * , , _ x _ r
A 9L\ _dL_ = XiAa + X2Ba + ... (16)
dt\dq a J dq a
CHAP. Ill
LAGRANGE'S EQUATIONS
285
It should be emphasized that the above results are applicable to holonomic (as well as
nonholonomic) systems since a constraint condition of the form
<l>{Ql,Q2, ...,Qn,t) = (17)
can by differentiation be written as
^pd Qa + &dt = (18)
which has the form (13).
(19)
LAGRANGE'S EQUATIONS WITH IMPULSIVE FORCES
Suppose that the forces F„ acting on a system are such that
lim C F v dt = Sv
where t represents a time interval. Then we call F„ impulsive forces and $ v are called
impulses.
If we let the subscripts 1 and 2 denote respectively quantities before and after appli
cation of the impulsive forces, Lagrange's equations become [see Problem 11.23]
'H.)  (2L) = f a
^dq a /2 \dq a /i
where
fa — 2* *Jv
dq a
(20)
(21)
If we call f a the generalized impulse, (20) can be written
Generalized impulse = change in generalized momentum
which is a generalization of Theorem 2.6, page 36.
(22)
Solved Problems
GENERALIZED COORDINATES AND TRANSFORMATION EQUATIONS
11.1. Give a set of generalized coordinates needed to completely specify the motion of
each of the following: (a) a particle constrained to move on an ellipse, (b) a circular
cylinder rolling down an inclined plane, (c) the two masses in a double pendulum
[Fig. 113] constrained to move in a plane.
(a) Let the ellipse be chosen in the xy plane of Fig. 111. The particle of mass m moving on the
ellipse has coordinates (x,y). However, since we have the transformation equations x = a cos 6,
y = 6 sin e, we can specify the motion completely by use of the generalized coordinate 0.
Fig. 111
Fig. 112
286 LAGRANGE'S EQUATIONS [CHAP. 11
(b) The position of the cylinder [Fig. 112 above] on the inclined plane can be completely specified
by giving the distance x traveled by the center of mass and the angle e of rotation turned
through by the cylinder about its axis.
If there is no slipping, x is related to e so that only one generalized coordinate [either x or e]
is needed. If there is slipping, two generalized coordinates x and e are needed.
(c) Two coordinates &i and 6 2 completely specify the positions of masses m l and w 2 [see Fig. 113
above] and can be considered as the required generalized coordinates.
11.2. Write the transformation equations for the system in Problem 11.1(c).
Choose an xy coordinate system as shown in Fig. 113. Let {x x ,y^) and {x^y^t be the rectangular
coordinates of m 1 and ra 2 respectively. Then from Fig. 113 we see that
x x = li cos $i 2/i = l\ sin $i
x 2 = h cos e i + l 2 cos 2 V2 ~ h sm * l + h sin 2
which are the required transformation equations.
11.3. Prove that t = ^.
dq<x dq a
We have r„ = r v (q u q 2 , ..., q n > t). Then
Thus * = £■ (•)
dq a dq a
We can look upon this result as a "cancellation of the dots".
11.4. Prove that ±(—) = !r
dt\dqj 3 #«
We have from (1) of Problem 11.3,
dr v d 2 r v . d 2 r„ . fl2r v
oq a dq a d<li d <la d <ln dq a dt
A (Oil)  JL { ?IL) !± 4 d ( dr v\ d( ln d ( ?>*v
N ° W dt\dqj ~ d qi \dq a J dt + '" dq n \d qoi J dt dt\dq a
d*r v . d 2r v . d2r„
' «i + " " * + a~ z~ «« + IT^T W
dtfl^« dq n dq<* dtdq a
Since r„ is assumed to have continuous second order partial derivatives, the order of differ
entiation does not matter. Thus from (2) and (3) the required result follows.
The result can be interpreted as an interchange of order of the operators, i.e.,
A(jL\ = — {A\
CLASSIFICATION OF MECHANICAL SYSTEMS
11.5. Classify each of the following according as they are (i) scleronomic or rheonomic,
(ii) holonomic or nonholonomic and (iii) conservative or nonconservative.
(a) A sphere rolling down from the top of a fixed sphere.
CHAP. Ill
LAGRANGE'S EQUATIONS 287
(b) A cylinder rolling without slipping down a rough inclined plane of angle a.
(c) A particle sliding down the inner surface, with coefficient of friction ^, of a
paraboloid of revolution having its axis vertical and vertex downward.
(d) A particle moving on a very long frictionless wire which rotates with constant
angular speed about a horizontal axis.
(a) scleronomic [equations do not involve time t explicitly]
nonholonomic [since rolling sphere leaves the fixed sphere at some point]
conservative [gravitational force acting is derivable from a potential]
(b) scleronomic
holonomic [equation of constraint is that of a line or plane]
conservative
(c) scleronomic
holonomic
nonconservative [since force due to friction is not derivable from a potential]
(d) rheonomic [constraint involves time t explicitly]
holonomic [equation of constraint is that of a line which involves t explicitly]
conservative
WORK, KINETIC ENERGY AND GENERALIZED FORCES
11.6. Derive equations (5) and (6), page 283, for the work done on a system of particles.
Suppose that a system undergoes increments dq lt dq 2 , . . . , dq n of the generalized coordinates.
Then the rth particle undergoes a displacement
" dr v
dr v = 2t dq a
a=l d 9«
Thus the total work done is
dW = 2 F„«dr v = 2 1 2 F,« — W?„ = 2 * a dq a
v=l v=l [a=l oq a j a =X
N dr
where * a — 2t F „ * t—
v=i oq a
We call * a the generalized force associated with the generalized coordinate q a .
11.7. Prove that <i> a = dW/dq a .
We have dW = ^j^dq a . Also, by Problem 11.6, dW = 2*«<*g a . Then
5 (♦.£)*. = »
Thus since the dq a are independent, all coefficients of dq a must be zero, so that * a = dW/dq a .
11.8. Let F„ be the net external force acting on the vth particle of a system. Prove that
d f^ . dr„~ ^ _ . dr v "V i? dTv
7u \Z, mvrv •  — y  > m v r„ • t— = 2/ F " * ^r
dt \v dq a ) v 5< Z« v oQct
By Newton's second law applied to the yth particle, we have
m„r„ = F„ (1)
288 LAGRANGE'S EQUATIONS [CHAP. 11
dr v dr„
Then m v r v >— = F„ • —  (2)
dq tt dq a
d /. 3r„\ .. dr v . d / $ r v
Now by Problem 11.4, rr ( r„ •  — ) = r„ •  — + r„ • — [  —
dt\ v dq a J " dq a v dt\dq a
_ v . ^ + ; . d ' v <*
dq a dq a
_,, .. 9t v d (• to v \ • &v m
dq a dt\ dq a / Bq a
Hence from (2) we have, since ra„ is constant,
d ( . dr„\ . 3r„ dr v
r: ( ra„ r„ •  — 1 — m v r„ •  — = F„ •  —
dt\ v " dq a J v dq a dq a
Summing both sides with respect to v over all particles, we have
d f« . 9r v \ v . 9r v « _ dr v
dt\^ m ^'^qJ^ m ^Iq a = ? F ^'^
11.9. Let T be the kinetic energy of a system of particles. Prove that
. . dT ^ . dic v ... dT ^ . dt v
v ' Bq a v dq a w dQa „ dq a
(a) The kinetic energy is T = r2%rj =  2 m v K ' K Thus
dT ^ . 9r v
—  2,m v r v ' —
dq a „ dq a
(b) We have by the "cancellation of the dots" [Problem 11.3, page 286],
dT v? • d *v V 3r "
3g a " dq a dq a
dr v
LAGRANGE'S EQUATIONS
11.10. Prove that ^ (£)  J^ = *«, « = 1, . . .,n where *«=F„ ^ .
From Problem 11.8,
d f« • 3r *l m • a *"  <? it . ar "
U)
From Problems 11.9(a) and 11.9(6),
¥ m » r »'3T a ~ W« {S)
Then substituting (2) and (3) in (1), we find
d(8T_)_f_ = „ a W
dt\dq a J dq a
(5)
The quantity Pa ~ J^~
is called the generalized momentum or conjugate momentum associated with the generalized coordi
nate q a .
CHAP. 11]
LAGRANGE'S EQUATIONS
289
11.11. Suppose that the forces acting on a system of particles are derivable from a potential
function V, i.e. suppose that the system is conservative. Prove that if LTV is
the Lagrangian function, then
dt\Bq a ) dq a
If the forces are derivable from a potential V, then [see Problem 11.7],
_ dW = dV
*« 3q a Bq a
Since the potential, or potential energy is a function of only the q's [and possibly the time t],
«L = 4<rv) = ^?
Bq a dq a 3q a
Then from Problem 11.10,
d /BL\ _ BT_ _ _ 3V_
dt \9q a ) 9q a dq a
d /3L\
dt\9qj
dL
=
11.12. (a) Set up the Lagrangian for a simple pendulum and
(b) obtain an equation describing its motion.
(a) Choose as generalized coordinate the angle e made by string
OB of the pendulum and the vertical OA [see Fig. 114].
If I is the length of OB, then the kinetic energy is
T = £mv 2 = %m{le)2 = £mP« 2 (1)
where m is the mass of the bob.
The potential energy of mass m [taking as reference
level a horizontal plane through the lowest point A] is given
by
V = mg(OA  OC) = mg(l — I cos e)
= mgl(l — cos e)
(«)
Thus the Lagrangian is
(b) Lagrange's equation is
From (3),
L = T
dL
Be
V = ^ml 2 e 2  mgl(l — cos e)
dt\Bi)
■ — mgl sin 9 ,
dL
Be
=
3L
Be
= ml 2 e
Substituting these in (4), we find
ml 2 % e + mgl sin e =
e + jrsin* =
(4)
(5)
which is the required equation of motion [compare Problem 4.23, page 102].
11.13. A mass Af 2 hangs at one end of a string which passes over a fixed frictionless non
rotating pulley [see Fig. 115 below]. At the other end of this string there is a
nonrotating pulley of mass Mi over which there is a string carrying masses mi and m%.
(a) Set up the Lagrangian of the system, (b) Find the acceleration of mass Af«.
Let X t and X 2 be the distances of masses M x and M 2 respectively below the center of the fixed
pulley. Let x x and x 2 be the distances of masses m x and m 2 respectively below the center of the
movable pulley M v
Since the strings are fixed in length,
X t + X 2 = constant = a, x x + x 2 = constant = 6
290
LAGRANGE'S EQUATIONS
[CHAP. 11
Then by differentiating with respect to time t,
X x + X 2 = or X 2 = —X x
and x x + x 2 = or x 2 = —x x
Thus we have
Velocity of M x = X x
Velocity of M 2 = X 2 = X x
d • •
Velocity of m x — j (X x + x x ) = X x + x x
Velocity of m 2 = — (X x + x 2 ) = X x + x 2 = X x — x x
Then the total kinetic energy of the system is Flg * 11_5
T = ±M X X 2 X + \M 2 ±\ + ±m x (X x + x x )2 + Lm 2 (k x x x )* (1)
The total potential energy of the system measured from a horizontal plane through the center
of the fixed pulley as reference is
V = M 1 gX l  M 2 gX 2  m x g(X x + x x )  m 2 g{X x + x 2 )
= M x gX x  M 2 fif(aZ])  m 1 g(X 1 + x 1 )  m 2 g(X 1 + b  x x ) (2)
Then the Lagrangian is
L = T  V
= %M X X.\ + \M 2 k\ + %m x (X x + x x )2 + \m 2 {X x  xtf
+ M x gX x + M 2 g{aX x ) + m 1 g(X x + x 1 ) + m 2 g(X x + b  x x ) (3)
Lagrange's equations corresponding to X x and x x are
Xj 3*1
_d fdL_\ _dL_ _
dt\dx x J dx x
U)
From (3) we have
jg = M x g  M 2 g + m x g + m 2 g
(M x  M 2 + m x + m 2 )g
dL •••#•• • •
—r — M X X X + M 2 X X + m x (X x + x x ) + m 2 (X x — x x ) = (M x + M 2 + m x + m 2 )X x + (m x — m 2 )x x
dX x
dL
— = m x g  m 2fl r = {m x m 2 )g
dL • • • . • .
— — = mi(X x + x x ) — m 2 {X x — x x ) = (m x — m 2 )X x + (m x + m 2 )x x
dXi
Thus equations (4) become
(M x + M 2 + m x + m 2 )X x + (m x — m 2 ) x x = (M x — M 2 + m x + m 2 )g
(m x — m 2 )X x + (m x + m^) x x = (m x — m 2 )g
Solving simultaneously, we find
X,
x x
(M x — M 2 )(m x + m 2 ) + 4m t m 2
(M x + M 2 )(m x + m 2 ) + 4m 1 m 2
2M 2 (m x — m 2 )
(M x + M 2 )(m x + m 2 ) + 4m x m 2
Then the downward acceleration of mass M 2 is constant and equal to
(M 2 — M x )(m x + m 2 ) — 4m x m 2
X 2 — —X x —
(M x + Af 2 )(m 1 + m 2 ) + 4m 1 ra 2
CHAP. Ill
LAGRANGE'S EQUATIONS
291
11.14. Use Lagrange's equations to set up the differential equation of the vibrating masses
of Problem 8.1, page 197.
Refer to Figs. 87 and 88 of page 197. The kinetic energy of the system is
T = \mx\ + \mx\ CO
Since the stretches of springs AP,PQ and QB of Fig. 88 are numerically equal to x lf x 2  x x and
x 2 respectively, the potential energy of the system is
V = %kx 2 1 + %k(x 2 x 1 ) 2 + &x 2 2 (*)
Thus the Lagrangian is
L = t  V = \mx\ + \mx\  \kx\  \k[x 2  xj*  &x\
Lagrange's equations are
d /BL\ SL _ d /dL\
dt{axj dx i ' dt \dxJ
dL
dx 9
=
Then since
dL
dL
= kx, + k(x 2  *i) = "(a?2  2x i)> T*~  mx i
dx, dx l
dL
dx 2
k{x 2 — * x ) — kx 2 = «(«! — 2x 2 ),
dL
dX 2
mx 2
equations (4) become m'x\ = k(x 2 2xJ, mx 2  k(x 1 2x 2 )
agreeing with those obtained in Problem 8.1, page 197.
(3)
(4)
(5)
11.15. Use Lagrange's equations to find the differential equation for a compound pendulum
which oscillates in a vertical plane about a fixed horizontal axis.
Let the plane of oscillation be represented by the xy plane
of Fig. 116, where O is its intersection with the axis of rota
tion and C is the center of mass.
Suppose that the mass of the pendulum is M, its moment
of inertia about the axis of rotation is I = MK 2 [K = radius
of gyration], and distance OC = h.
If e is the instantaneous angle which OC makes with^the
vertical axis through O, then the kinetic energy is T = %I 6 2 =
±MK 2 2 . The potential energy relative to a horizontal plane
through is V = —Mgh cos 6. Then the Lagrangian is
L = T V = \MK 2 e 2 + Mgh cos e
Since dL/do  Mgh sin e and dLIdd  MK 2 e, Lagrange's
equation is
d_ / dL
dt[do
dL
de
i.e.,
MK 2 6 + Mgh sin = or
Compare Problem 9.24, page 237.
V + ^ sin « =
Fig. 116
11.16. A particle of mass m moves in a conservative force field. Find (a) the Lagrangian
function, (b) the equations of motion in cylindrical coordinates ( P , 4>, z) [see Problem
1.147, page 32].
(a) The total kinetic energy T  %m[p 2 + p 2 $ 2 + z 2 } . The potential energy V = V(p,<p,z). Then
the Lagrangian function is
L = TV = lm[p2 + p 2^2 + 2] V( p ,0,s)
292 LAGRANGE'S EQUATIONS [CHAP. 11
(b) Lagrange's equations are
d /8L\ dL _ . d. • / .„ dV\ .. . dV
d /8L\ dL _ . & 2 \,dV n d , „v dV
d fdL\ dL n . d..,av n .. d v
11.17. Work Problem 11.16 if the particle moves in the xy plane and if the potential
depends only on the distance from the origin.
In this case V depends only on P and z = 0. Then Lagrange's equations in part (6) of Problem
11.16 become
rnGp'^) = ^, A( P 20) = o
These are the equations for motion in a central force field obtained in Problem 5.3, page 122.
LAGRANGE'S EQUATIONS FOR NONHOLONOMIC SYSTEMS
11.18. Derive Lagrange's equations (15), page 284, for nonholonomic constraints.
Assume that there are m constraint conditions of the form
2 A a dq a + Adt = 0, 2 B a dq a + B dt = 0, ... (1)
a a
where m < n, the number of coordinates q a .
As in Problem 11.10, page 288, we have
v _ d / dT \ 8T x .. dr v
If Sr„ are virtual displacements which satisfy the instantaneous constraints [obtained by consider
ing that time t is a constant], then
^ dr v
Sr„ = 2^5g a (3)
Now the virtual work done is
dr
SW = 2 «**¥„ • 8r v = 2 2 m„ r„ • —  S? a = 2 Y a 8q a (4)
v v a oq a a
Now since the virtual work can be written in terms of the generalized forces <t> a as
SW = 2* a S 9a (5)
a
we have by subtraction of (4) and (5),
2(^a*a)Sg a = (6)
a
Since the 8q a are not all independent, we cannot conclude that Y a = $ a which would lead to
Lagrange's equations as obtained in Problem 11.10.
Prom (1), since t is constant for instantaneous con train ts, we have the m equations
J,A a 8q a = 0, lB a Sq a = 0, ... (7)
a a
Multiplying these by the m Lagrange multipliers X lt X 2 , ... and adding, we have
2 (M a + X 2 B a + • • •) 8q a = (8)
CHAP. 11]
LAGRANGE'S EQUATIONS
293
Subtraction of (6) and (8) yields
2 (Y a  *«  M B  X 2 B a ) 8q a =
(9)
Now because of equations (7) we can solve for m of the quantities 8q a [say 5^, . . .,8q m ] in terms
of the remaining 8q a [say 8g m + 1 , • • ., 8q n ]. Thus in (9) we can consider 8 ffl> . . ., S<? m as dependent
and Sg m + i, . . ., 8q n as independent.
Let us arbitrarily set the coefficients of the dependent variables equal to zero, i.e.,
(10)
*a — X l^a ~ X 2 B a
= o,
1,2, ...,m
Then there will be left in the sum (9) only the independent quantities 8q a and since these are ar
bitrary it follows that their coefficients will be zero. Thus
Equations (2), (10) and (11) thus lead to
= 0,
m+1, . . .,«
(«)
d/3T\_T = * a+Xl A a + X a F a + ••■ « = l,2,...,n
as required. These equations together with (i) lead to n + m equations in n + m unknowns.
11.19. Derive equations (16), page 284, for conservative nonholonomic systems.
From Problem 11.18,
d /dT
dT
= <f> a + X x A a + X 2 B« +
(1)
dt \dq a J d <la
Then if the forces are derivable from a potential, <t>« = dV/dq a where V does not depend on
q a . Thus (1) can be written
where L = T — V.
dt\dq a J d <la
11.20. A particle of mass m moves under the influence
of gravity on the inner surface of the paraboloid
of revolution x 2 + y 2 = az which is assumed
frictionless [see Fig. 117]. Obtain the equations
of motion.
By Problem 11.16, the Lagrangian in cylindrical co
ordinates is given by
L = ±m('p 2 + P 2 2 + I 2 )  mgz
{D
Since x 2 + y 2 = p 2 , the constraint condition is p 2 — az =
so that
2p 8p — a 8z
(2)
If we call Qi = p, q% = <t>, <lz — z and compare (2) with
the equations (7) of Problem 11.18, we see that
Fig. 117
A x = 2p, A 2 = 0, A 3 = a (3)
since only one constraint is given. Lagrange's equations [see Problem 11.19] can thus be written
d /1L\ *k  xA
1,e " <tt\dp) 3p
Using (1), these become
2X lP ,
a = 1,2, 3
<ft^4
d^
o,
dt\ dz
dz
XjO.
294 LAGRANGE'S EQUATIONS [CHAP. 11
m('pp$ 2 ) = 2X lP (4)
m'z = —mg — \ t a (6)
We also have the constraint condition
2pp — az = (7)
The four equations &), (5), (6) and (7) enable us to find the four unknowns p, $, z, \ v
11.21. (a) Prove that the particle of Problem 11.20 will describe a horizontal circle in the
plane z = h provided that it is given an angular velocity whose magnitude is
(o = y/2g/a .
(b) Prove that if the particle is displaced slightly from this circular path it will
undergo oscillations about the path with frequency given by (lh)^2g/a .
(c) Discuss the stability of the particle in the circular path.
(a) The radius of the circle obtained as the intersection of the plane z = h with the paraboloid
p 2 = az is . —
Po = Va>h (1)
Letting z = h in equation (6) of Problem 11.20, we find
Xi = —mg/a {2)
Then using (1) and (*) in equation (4) of Problem 11.20 and calling = w, we find m( Po u 2 ) =
2{—mg/a)p Q or w 2 = 2g/a from which
<o = V2g/a (3)
The period and frequency of the particle in this circular path are given respectively by
P 1 = 2,^ and /,=£>/? «»
(b) From equation (5) of Problem 11.20, we find
p 2 = constant = A (5)
Assuming that the particle starts with angular speed w, we find A = aha so that
^ = ahu/p 2 (6)
Since the vibration takes place very nearly in the plane z — h, we find by letting z = h
in equation (6) of Problem 11.20 that
\j = —mg/a (7)
Using (6) and (7) in equation (4) of Problem 11.20, we find
P  a 2 h 2 a 2 /p 3 = 2£p/a (8)
Now if the path departs slightly from the circle, then p will depart slightly from p . Thus we
are led to make the transformation
p = po + u (9)
in (8), where u is small compared with p . Then (8) becomes
(p + u) 6 a
But to a high degree of approximation,
11 _ i/ 1+ iiV 3 = l/i_ 3w
(p + tt) 3 p3(l+w/p )3 p3\ p o y p 3\ p
by the binomial theorem, where we have neglected terms involving u 2 , u s , . . . . Using the
values of p and a given by (1) and (5) respectively, (10) becomes
CHAP. 11] LAGRANGE'S EQUATIONS 295
u 4 (Sg/a)u = ( 5 )
whose solution is it = e t cos V8«r/o t + e 2 sin VSg/a t. Thus
p = Po + u = Vah + ej cos VSgJa t + e 2 sin y/Sg/a t
It follows that if the particle is displaced slightly from the circular path of radius Po = Vah,
it will undergo oscillations about the path with frequency
t>  \~°L (7)
or period 2 ~ v "\2g
It is interesting that the period of oscillation in the circular path given by (4) is twice
the period of oscillation about the circular path given by (7).
(c) Since the particle tends to return to the circular path when it is displaced slightly from it,
the motion is one of stability.
11.22. Discuss the physical significance of the Lagrange multipliers Xi, A 2 , . . . in Problem
11.18.
In case there are no constraints the equations of motion are by Problem 11.10,
±/dT\ _ 9T_ =
dt \dq a ) d Qa
In case there are constraints the equations are by Problem 11.18,
dt \dq a J d< *«
It follows that the terms \ x A a + \ 2 B a + • • • correspond to the generalized forces associated with
constraints.
Physically, the Lagrange multipliers are associated with the constraint forces acting on the
system. Thus when we determine the Lagrange multipliers we are essentially taking into account
the effect of the constraint forces without actually finding these forces explicitly.
LAGRANGE'S EQUATIONS WITH IMPULSIVE FORCES
11.23. Derive the equations (20), page 285.
For the case where forces are finite we have by Problem 11.10,
dt{ dq J dq a
where *« = 2 F„ • t— ( g )
v °*ia
Integrating both sides of (1) with respect to t from t = to t = r,
Jo dt \*Qa) Jo a< *« J
(5)
so that ( —7 ) —
\dq a /t=T
Taking the limit as r *■ 0, we have
lim
lim f*Ldt = 2 (Aim C *,*)>£
auSL(SL>»rs*  n*s;'*)'Z
296
LAGRANGE'S EQUATIONS
[CHAP. 11
f) (f) = JA^ = T.
a — dt = since  — is finite, and lim I F v dt = tf„.
«i °tfa »9o t»0 J„
Fig. 118
11.24. A square ABCD formed by four rods of length
21 and mass m hinged at their ends, rests on a
horizontal frictionless table. An impulse of
magnitude 3 is applied to the vertex A in the
direction AD. Find the equations of motion.
After the square is struck, its shape will in gen
eral be a rhombus [Fig. 118].
Suppose that at any time t the angles made by
sides AD (or BC) and AB (or CD) with the x axis
are e t and 6 2 respectively, while the coordinates of
the center M are (x, y). Thus x,y,0 x ,6 2 are the gen
eralized coordinates.
From Fig. 118 we see that the position vectors
of the centers E,F,G,H of the rods are given re
spectively by
r E = (x — I cos 0{)i + (y — I sin e t )j
r F = (x + I cos 6 2 )\ + (y — I sin e 2 )j
r G = (x + I cos 0i)i + (y + I sin e^j
r H = {x — I cos e 2 )i + (y + I sin 6 2 )j
The velocities of E, F, G and H at any time are given by
v E = r E = (x + I sin X eji + (y — I cos $i x )j
v F = r F = (x — I sin e 2 $ 2 )i + (y — I cos o 2 e 2 )j
v g = *g — {x — I sin $! #i)i + (y + I cos e x b i)j
v H = r H = (x + I sin 2 #2)* + (if + ^ co . s *2 *2)j
The kinetic energy of a rod such as AZ? is the same as the kinetic energy of a particle of
mass m located at its center of mass E plus the kinetic energy of rotation about an axis through
E perpendicular to the xy plane. Since the angular velocity has magnitude e 2 aR d the moment of
inertia of a rod of length 21 about its center of mass is I AB = $ml 2 , the total energy of rod AB is
Similarly, the total kinetic energies of rods BC, CD and AD are
T BC = %mrj. + lI BC el, T CD = %mr% + $I CD »l, T AD = %mi 2 H + %I AD o\
Thus the total kinetic energy is [using the fact that I = I AB = I BC = Icd = £ m * 2 ]
T — T AB + T BC + T CD + T AD
= m(r + r F + r% + r 2 H ) + I(e\ + «f )
= im(4^2 + 4y 2 + 2Z 2 * 2 + 2J2* 2 .) + £ mZ 2(^ + ^
= 2m(x 2 + y 2 ) + %ml 2 {e\ + e 2 2 )
Let us assume that initially the rhombus is a square at rest with its sides parallel to the
coordinate axes and its center located at the origin. Then we have
x = 0, y = 0, <?! = v/2, e 2 = 0, x = 0, y = 0, ^ = 0, e 2 =
If we use the notation ( H and ( ) 2 to denote quantities before and after the impulse is applied, we
have „.
'2Z\ = (4«i)i = ( " ) = (4my)i =
Bx/x \ d V/i
CHAP. 11] LAGRANGE'S EQUATIONS 297
dT \ / ST \
— r ) = (Amx) 2 = Am'x ( — r ) = (Amy) 2 = 4m£
3*/ 2 V%/ 2
(2)
(S)
04)
(5)
\™l/2
/dT
\de 2
• ) = §raZ 2 2 = \mVh
/2
(5).®. = '■
or
4m* = f x
(fe) 2 "(fe)i = ftf
or
Amy  f y
(i?),"(®i = ^
or
JmPff! = y 6i
{'df 2 ) 2 ~\dfj 1 = Te *
or
fmPffg = fe 2
where for simplicity we have now removed the subscript ( ) 2 .
To find f x , f y , f $i , f h we note that
T* =
2j„
V
where ^ are the impulsive forces. We thus have
dr A dr B dr r dr n
dr A 3r B dr c dr D
T. = Sa£+ A^ + A^+ A, •■£ (T)
3r A dr R 3r c 3r D
3r A 3r B 9r c dr D
Now from Fig. 118 we find the position vectors of A,B,C,D given by
r A = (x — I cos 9 1 — I cos 2 )i + (2/ ~ I sin x + I sin 2 )j
r B = (x — I cos tf ! + I cos 2 )i + (2/ — J sin e 1 — I sin fl 2 )j
r c = (x + I cos x + I cos e 2 )i + (y + I sin 8 X — I sin $ 2 )j
td = (x + I cos e x — I cos e 2 )i + (y + I sm$i + I sin 2 )J
Since the impulsive force at A is initially in the direction of the positive y axis, we have .
c£a = Si do)
Thus equations (6)(9) yield
fx = 0, 7^ = J, y 9l =  J* cos e x , Te 2 = SI cos 6 2 (11)
Then equations (1)(U) become
4m* = 0, Amy = J, fm/ 2 ^ = — SI cos tf^ mZ 2 ff 2 = £1 cos 2 (Z2)
11.25. Prove that the kinetic energy developed immediately after application of the impulsive
forces in Problem 11.24 is T = S 2 /2m.
From equations (12) of Problem 11.24, we have
s %s * %s
* = °' V = 4m> * = ~8mT C0S ^' '* = 8m7 cos '*
298
LAGRANGE'S EQUATIONS
[CHAP. 11
Substituting these values in the kinetic energy obtained in Problem 11.24, we find
<72 3 (12
But immediately after application of the impulsive forces, e x = tt/2 and 6 2 = approximately.
Thus (1) becomes T = J 2 /2m.
MISCELLANEOUS PROBLEMS
11.26. In Fig. 119, AB is a straight frictionless wire
fixed at point A on a vertical axis OA such that
AB rotates about OA with constant angular
velocity <■>. A bead of mass m is constrained to
move on the wire, (a) Set up the Lagrangian.
(b) Write Lagrange's equations, (c) Determine
the motion at any time.
(a) Let r be the distance of the bead from point A of
the wire at time *. The rectangular coordinates
of the bead are then given by
x = r sin a cos at
y = r sin a sin at
z — h — r cos a
where it is assumed that at t — the wire is in
the xz plane and that the distance from O to A is h.
The kinetic energy of the bead is
T = ±m(x 2 + y 2 + z 2 )
— ±m{(r sin a cos at — ar sin a sin at) 2
Fig. 119
+ (r sin a sin at + ar sin a cos at) 2 + (— r cos a) 2 }
= ^m(r 2 + co 2 r 2 sin 2 a)
The potential energy, taking the xy plane as reference level, is V = mgz = mgr(/i  r cos a).
Then the Lagrangian is
L = T — V — \m(r 2 + a 2 r 2 sin 2 a)  mg(h — r cos a)
(b) We have
flL
— = m« 2 r sin 2 a + mg cos a,
dr
§L_
dr
= mr
and Lagrange's equation is ^ ( ^
dL
dr
— or
i.e.,
mr — (ma 2 r sin 2 a + w# cos a) —
r — (w 2 sin 2 a)r = fif cos a
CO
(c) The general solution of equation (1) with the right hand side replaced by zero is
c e (a sin a)t f C2 e (co sin a)t
Since the right hand side of (1) is a constant, a particular solution is ^^. Inus tne
<o z sin z a
general solution of (1) is
fif cos a
r = ^CasinaOt + C2 eCa, sin aK _ __ —
to 2 sin z a
This result can also be written in terms of hyperbolic functions as
r = c 3 cosh (« sin a)i + «4 sinh (« sin a)t
g cos a
to 2 sin 2 a
(2)
(3)
CHAP. 11] LAGRANGE'S EQUATIONS 299
11.27. Suppose that in Problem 11.26 the bead starts from rest at A. How long will it take
to reach the end B of the wire assuming that the length of the wire is It
Since the bead starts from rest at t = 0, we have r = 0, r = at t = 0. Then from equa
tion (2) of Problem 11.26,
, g cos a , n
Ci + c 2 = o ■ o and c x — c 2 =
1 * w 2 sin 2 a
Thus c x = Co = ^ — ttt and (2) of Problem 11.26 becomes
1 2w 2 sm 2 a v '
„ _ _JL£2!L£_ / e (<a sin a)t J e (u sin a)t\ — & C0S a /*\
2« 2 sin 2 a
flr COS a , , , . . ....
■4 — r^— {cosh (w sm a)£ — 1}
w 4 Sir a
which can also be obtained from equation (3) of Problem 11.26. When r = Z, (2) yields
Z« 2 sin 2 a
so that the required time is
cosh (« sin a)* = 14
9 cos a
* = 4coshVlf Z " 2si f a N )
w sm a \ flr cos 2 a /
= dn^ ln {( 1+ ^^UA/fl + ^^) 1
g COS 2 a / V I gr cos 2 a
11.28. A double pendulum [see Problem 11.1(c) and Fig. 113, page 285] vibrates in a
vertical plane, (a) Write the Lagrangian of the system, (b) Obtain equations for
the motion.
(a) The transformation equations given in Problem 11.2, page 286,
x x = l x cos 9 X y x = l x sin $ x
x 2 = l x cos X 4 Z 2 cos 6 2 V2 — h sin *i 4 l 2 sin e 2
yield »! = l x e x sin ^ ^ = l x e x cos ^
»2 = ~h 9 \ sin tf x — Z 2 2 sin e 2 y 2 = Z^ cos e x 4 l 2 e 2 cos e 2
The kinetic energy of the system is
T = $m x {x\ + y 2 x ) + ^m&l + fy
= %m x l\e\ 4 \m 2 \$o\ 4 Z^a + 2Z 1 Z 2 * 1 * 2 cos (^  * 2 )]
The potential energy of the system [taking as reference level a plane at distance l t + l 2
below the point of suspension of Fig. 113] is
V = m x g[l x 4 Z 2  l x cos e x ] + m 2 g[l x 4 Z 2  (Z x cos 0! + Z 2 cos 2 )]
Then the Lagrangian is
L = T  V
= \™>xA<>\ + %™>2{l 2 j\ + * + 2Z 1 Z 2 m 2 cos (e x  e 2 )) (i)
m x g[l x + l 2 l x cos $ x ]  m2g[l x + l 2  (l x cos e x 4 Z 2 cos 6 2 )]
(b) The Lagrange equations associated with e x and 2 are
300 LAGRANGE'S EQUATIONS [CHAP. 11
From (1) we find
dL/de x — —m 2 l x l 2 e x e 2 sm ( e i ~ e z) ~ m idh sin e \ ~ m 20h sin e x
dL/de x = ml\e x + m 2 l\e x + m 2 l x l 2 e 2 c °s (<?i — o 2 )
dL/de 2 = m 2 l x l 2 e x e 2 sin (*i — 2) ~ m*zQh sin 2
dL/de 2 = m 2 l\e 2 + m 2 l x l 2 9 x cos {e x — e 2 )
Thus equations {2) become
m x f x e\ + m 2 l\ e\ + m 2 l x l 2 'e 2 cos (*i ~ #2) ~ rn 2 l x l 2 e 2 (e x — e 2 ) sin (e x — 2 )
 m 2 l x l 2 6 x 6 2 sin (e x — e 2 ) — m x gl x sin e x — m 2 gl x sin e x
and w 2 ?2 * 2 + m 2 l x l 2 «i cos (<?!  e 2 )  m 2 l x l 2 e\(e x  e 2 ) sin {e x  o 2 )
= m 2 l x l 2 x e 2 sin (<?! — 9 2 ) — m 2 gl 2 sin <9 2
which reduce respectively to
(m x + m 2 )l\ e\ + m 2 l x l 2 e 2 cos (<?!  e 2 ) + m 2 l x l 2 e\ sin (e x  e 2 ) = {m x + m 2 )gl x sin e x (3)
and m 2 Z 2 *0 2 + m 2 l x l 2 'e\ cos (0 x 6 2 )  m 2 l x l 2 o\ sm(o x e 2 ) = m 2 gl 2 sine 2 (4)
11.29. Write the equations of Problem 11.28 for the case m x = m 2 = m and h = h = I
Letting m y = m 2 , l x = Z 2 in equations (3) and (4) of Problem 11.28 and simplifying, they
can be written
21 e\ + I e 2 cos (e x  e 2 ) + lei sin {e x  e 2 ) = 2g sin e x {1)
I 'e x cos (e x  e 2 ) + I 'e 2  le\ sin {e x  e 2 ) = g sin e 2 (2)
11.30. Obtain the equations of Problem 11.29 for the case where the oscillations are assumed
to be small.
Using the approximations sin e = 6, cos e = 1 and neglecting terms involving 6 2 e, the equa
tions (1) and (2) of Problem 11.29 become
2l'e x + l'e 2  2ge x
l'e\ + l'o 2 = —go 2
11.31. Find the (a) normal frequencies and (b) normal modes corresponding to the small
oscillations of the double pendulum.
(a) Let e x = A x cos*t, * 2 = A 2 cos<ot [or A x e^, A 2 e^] in the equations of Problem 11.30.
Then they can be written
2(glo> 2 )A x  1<* 2 A 2 = 0
lo> 2 A x + (gl<* 2 )A 2 = Oj CO
In order for A x and A 2 to be different from zero, we must have the determinant of the coefficients
equal to zero, i.e.,
2{g  Zo) 2 ) Zco 2
hfl g ~ to 2
=
or Z 2 co 4  4tow 2 + 2g 2 = 0. Solving, we find
4lg ± y/l6l 2 g 2  M 2 9 2 (2±V^)9
,.& =
2l 2 I
{2 + y/2)g 2 _ ( 2  y[2)g
A = — 1 — ' W2  I
CHAP. Ill LAGRANGE'S EQUATIONS 301
Thus the normal frequencies are given by
Wl 1 l(2 + V2)g _ tt2 _ i J (2V2)g
A = 2^ = 2^\ I and ft ~ tor ~ & \ I
w
(b) Substituting a 2 = a* = (2 + y/2)g/l in equations (1) of Part (a) yields
A 2 = V2A t U)
This corresponds to the normal mode in which the bobs are moving in opposite directions.
Substituting u 2 = w 2 = (2 — \/2 )gll in equations (1) of Part (a) yields
A 2 = V2A! (5)
This corresponds to the normal mode in which the bobs are moving in the same directions.
11.32. (a) Set up the Lagrangian for the motion of a symmetrical top [see Problem 10.25,
page 268] and (b) obtain the equations of motion.
(a) The kinetic energy in terms of the Euler angles [see Problem 10.24, page 268] is
T = \{h»\ + 1 2 4 + h4) = %h& sin2 e + g2) + y^ cos e + fo (i)
The potential energy is V = mgl cos 8 (2)
as seen from Fig. 1018, page 269, since distance OC = I and the height of the center of
mass C above the xy plane is therefore I cos 8. Thus
L = T  V = J/^ 2 sin2 e + £2) + / 3 (0 cos e + J)2 _ mflr ; C os (5)
(6) 3L/d0 = i^ 2 sin 8 cos + / 3 (<£ cos 8 + ^){—<t> sin 8) + mflfZ sin *
dL/dd = J^
3L/30 =
dL/d$ — Irf sin 2 8 + I s ($ cos + 4>) cos
3L/d^ =
dL/dj, — 7 3 (0 cos 8 + ^)
Then Lagrange's equations are
±/9L\_9L  n A/^.\_^ = n <L/i>LL\ Ok 
or I\0 ~ h4> 2 sin cos + / 3 (^ cos + \p)$ sin — mflrZ sin = (4)
— [7 t sin 2 6 + J 3 (0 cos 8 + $) cos 0] = (5)
j^[/ 3 (^cos0 + ^)] = (6)
11.33. Use the results of Problem 11.32 to obtain agreement with the equation of
(a) Problem 10.29(&), page 270, and (b) Problem 10.27(a), page 270.
(a) From equations (5) and (6) of Problem 11.32 we obtain on integrating,
Itf sin 2 8 + I 3 ($ cos 8 + ^) cos 8 = constant = K (1)
<£ cos * + ^ = A (2)
Using (2) in (1), we find IJj> sin 2 + 7 3 A cos = J£
(6) Using (2) in equation (4) of Problem 11.32, we find
ij o — Ji<£ 2 sin 8 cos + / 3 A<£ sin 8 = mgrZ sin 8
302
LAGRANGE'S EQUATIONS
[CHAP. 11
11.34. Derive Euler's equations of motion for a rigid body by use of Lagrange's equations.
The kinetic energy in terms of the Euler angles is [see Problem 10.24, page 268]
T = £(J lW ? + I 2 4 + 7 8 „i)
 J/^ sin <? sin ,/< + 6 cos \J) 2 + £/ 2 (<£ sin e cos ^  6 sin ^) 2 + %h(4> cos + «/<) 2
Then dT/dxp  I X Q> sin sin ^ + (9 cos </<)(0 sin e cos^p — e sin ./>)
+ I 2 (<£ si n * cos \p — 6 sin ^)(— <£ sin sin >/< — cos ^)
= / 1 w 1 w 2 + J 2 (w 2 )(— a^) = (^1 — ^2) w l w 2
ar/af = i 3 ($ cos + «/>) = ?3 w 3
Then by Problem 11.10, page 288, Lagrange's equation corresponding to ^ is
d_/dT\ 9T _
or ^3"3 + (h ~ ^i) w i"2 = *</; (■*)
This is Euler's third equation of (22), page 256. The quantity *^ represents the general
ized force corresponding to a rotation ^ about an axis and physically represents the component
A 3 of the torque about this axis [see Problem 11.102].
The remaining equations , . /T t ^ . ,„\
7 lWl + (I 3  i 2 ) w 2 w 3 = A l \ 2 )
J 2 « 2 + (h  ^3)"3«l = A 2 W
can be obtained from symmetry considerations by permutation of the indices. They are not directly
obtained by using the Lagrange equations corresponding to and <£ but can indirectly be deduced
from them [see Problem 11.79].
11.35. A bead slides without friction on a f rictionless
wire in the shape of a cycloid [Fig. 1110]
with equations
x = a(0 — sin 9), y = a(l + cos 9)
where ^ 9 ^ 2tt. Find (a) the Lagrangian
function, (b) the equation of motion.
(a) Kinetic energy = T = m(x 2 + y 2 )
= lma 2 {[(l  cos e)d] 2 + [sin e] 2 }
— ma 2 (l — cos e)e 2
Potential energy = V  mgy — mga{l + cos 0)
Then
Lagrangian = L = T V = ma 2 (l  cos 0)0 2  mga(l + cos 0)
Fig. 1110
... d fdL\ dL n .
d
[2ma 2 (l — cos 0)0] — [ma 2 sin 6 e 2 + mga sin e] =
9
dt
[(1  cos 9)e]  \ sin 6 2  — sin 6 =
which can be written (1  cos 6) e + % sin 6 e 2  g sin 6 
11.36. (a) Show that the equation of motion obtained in part (b) of Problem 11.35 can be
written , 2
a u + SL W = o where w = cos (5/2)
dt 2
4a
CHAP. 11] LAGRANGE'S EQUATIONS 303
and thus (b) show that the bead oscillates with period 2 7 r\ / 4a/g .
(a) If u = cos (0/2), then
~r = ± sin (0/2)0, ^ = A sin (0/2)0* A cos (0/2)02
Thus vg + ju = is the same as
\ sin (0/2) V  i cos (0/2)02 + ■■& cos (0/2) =
+ i cot (0/2)0*2  L cot (0/2) = (i)
2 —'' 4~~v,/ ■ 4a
which can be written as
Since cot (0/2) = ^ os (<?/2) = 2 sin (0/2) cos (0/2) sin
w ' sin (0/2) 2 sin2 (0/2) 1  cos
it follows that equation (1) is the same as that obtained in Problem 11.35(&).
(b) The solution of the equation is
u = cos (0/2) = Cj cos V4a/flr t + c 2 sin y/ba/g t
from which we see that cos (0/2) returns to its original value after a time 2iry/£a/g which
is the required period. Note that this period is the same as that of a simple pendulum with
length I — 4a.
An application of this is the cycloidal pendulum. See Problem 4.86, page 112.
11.37. Obtain equations for the rolling sphere of Problem 9.42, page 244 by use of Lagrange's
equations.
Refer to Fig. 933 in which <t> and ^ represent generalized coordinates. Since the sphere of
radius CP = a rolls without slipping on the sphere of radius OP = b, we have
6 d<p/dt — a d\f//dt or b$ = af
which shows that if <p = when ^ — 0, then
b<j> = a\p (i)
Thus </> and $ [and therefore d<p and d> or 8<t> and 3^] are not independent.
The kinetic energy of the rolling sphere is
T = im(a + 6)2^2 + _ij w 2
= ±m(a + 6)2^2 + ^(ma2 )( ; + ^2
using the fact that I = %ma? is the moment of inertia of the sphere about a horizontal axis
through its center of mass.
The potential energy of the rolling sphere [taking the horizontal plane through O as reference
level] is
V = mg(a + b) cos <f>
Thus the Lagrangian is
L = T  V = {m(o + 6) 2 02 + ±ma 2 (4> + J)2  m g(a + b) cos <f> (2)
We use Lagrange's equations (16), page 284, for nonholonomic systems. From (1) we have
b 8<f> — a S^ = (#\
so that if we call g x = and q 2  f and compare with equation (7) of Problem 11.18, page 292
we find '
A t = b, A 2 = —a (4)
Thus equations (16), page 284, become
d L /dL\ dL _
_d /BL\ dL _
304 LAGRANGE'S EQUATIONS [CHAP. 11
Substitution of (2) intd (5) and (6) yields
m(a+b) 2 $ + fma 2 (0 + £)  mg(a + b) sin <f>  \ x b (7)
fraa 2 (0 + 'vfr) = X x a (8)
Substituting ^ = (b/a)</> [from (1)] into (7) and (8), we find
m(a + b) 2 5? + §raa 2 (l + 6/a) £  wfir(o + 6) sin <p = \ x b (9)
. . fma 2 (l + 6/a)0 = XjO (10)
Now from (10) we have Xi = —%m(a + b)$
and using this in (9) it becomes after simplifying and solving for *<p,
•• 5flf
* ~ 7(^+6) 8m *
This is the same equation as that of (2) in Problem 9.42, page 244, with <f>  w/2  e. To find the
required angle at which the sphere falls off, see Problem 11.104.
11.38. (a) Solve the equations of motion obtained in Problem 11.24, page 296, and (b) inter
pret physically.
(a) From the first of equations (12) in Problem 11.24 we have
<c = constant = (0
since x = at t = 0. Similarly, from the second of equations (12) we have
* = is' «
since y — at t = 0.
From the third of equations (12) we find on separating the variables,
or on integrating, In cot I ^  — J  ~g^J + c i
i.e.
tan(^) = c 2 eW* ml
Thus since X = n/2 at < = 0, we have c 2 = 0. This means that for all time we must
have $i = n72.
From the fourth of equations (12) in Problem 11.24 we have similarly,
sec 9 2 de 2  z^jdt
e 2 \ 3Jt
or on integrating, In cot (7 — — J  ^—f + c 3
i.e., tan (i"^) =' W~ ZSmml
Now when t = 0, e % = so that c 4 = 1. Then
ten/EM = e 8^8«i or , 2 =  2 tani( e  3 ^ /8w
(6) Equations (1) and (2) show that the center moves along the y axis with constant speed J/4m.
The rods AD and BC are always parallel to the y axis while rods AB and CD slowly rotate until
finally [*»"] the rhombus collapses, so that all four rods will be on the y axis.
CHAP. 11] LAGRANGE'S EQUATIONS 305
Supplementary Problems
GENERALIZED COORDINATES AND TRANSFORMATION EQUATIONS
11.39. Give a set of generalized coordinates needed to completely specify the motion of each of the follow
ing: (a) a bead constrained to move on a circular wire; (6) a particle constrained to move on a
sphere; (c) a compound pendulum [see page 228]; (d) an Atwood's machine [see Problem 3.22,
page 76]; (e) a circular disk rolling on a horizontal plane; (/) a cone rolling on a horizontal plane.
11.40. Write transformation equations for the motion of a triple pendulum in terms 'of a suitable set of
generalized coordinates.
11.41. A particle moves on the upper surface of a frictionless paraboloid of revolution whose equation
is x 2 + y 2 = cz. Write transformation equations for the motion of the particle in terms of a suit
able set of generalized coordinates.
11.42. Write transformation equations for the motion of a particle constrained to move on a sphere.
CLASSIFICATION OF MECHANICAL SYSTEMS
11.43. Classify each of the following according as they are (i) scleronomic or rheonomic, (ii) holonomic
or nonholonomic, and (iii) conservative or nonconservative:
(a) a horizontal cylinder of radius a rolling inside a perfectly rough hollow horizontal cylinder of
radius b > a;
(6) a cylinder rolling [and possibly sliding] down an inclined plane of angle a;
(c) a sphere rolling down another sphere which is rolling with uniform speed along a horizontal
plane;
(d) a particle constrained to move along a line under the influence of a force which is inversely
proportional to the square of its distance from a fixed point and a damping force proportional
to the square of the instantaneous speed.
Ans. (a) scleronomic, holonomic, conservative
(b) scleronomic, nonholonomic, conservative
(c) rheonomic, nonholonomic, conservative
(d) scleronomic, holonomic, nonconservative
WORK, KINETIC ENERGY AND GENERALIZED FORCES
11.44. Prove that if the transformation equations are given by r v = r v (q v q 2 , . . ., q n ), i.e. do not involve
the time t explicitly, then the kinetic energy can be written as
n n
T = 2 2 a a a la 9b
where a a/3 are functions of the q a .
11.45. Discuss Problem 11.44 in case the transformation equations depend explicitly on the time t.
11.46. If F(Xx, Xy, Xz) = X n F(x, y, z) where X is a parameter, then F is said to be a homogeneous function
of order n. Determine which (if any) of the following functions are homogeneous, giving the order
in each case:
(a) x 2 + 2/ 2 + z 2 + xy + yz + xz (e) x s tan 1 (y/x)
(6) 3x2y + 4z (/) 4 sin xy
(c) xyz + 2xy + 2xz + 2yz (g) (x + y + z)/(x 2 + y 2 + z 2 )
(d) (x + y + z)/x
Ans. (a) homogeneous of order 2, (6) homogeneous of order 1, (c) nonhomogeneous, (d) homo
geneous of order zero, (e) homogeneous or order 3, (/) nonhomogeneous, (g) homogeneous of
order —1.
11.47. If F{x,y,z) is homogeneous of order n [see Problem 11.46], prove that
dF , dF , dF
a— + yz h 2t = nF
dx " by dz
This is called Euler's theorem on homogeneous functions.
[Hint. Differentiate both sides of the identity F(\x,\y,\z)  X n F(x,y,z) with respect to X and
then place X = 1.]
11.48. Generalize the result of Problem 11.47.
306 LAGRANGE'S EQUATIONS [CHAP. 11
11.49. Prove that if the transformation equations do not depend explicitly on time t, and T is the kinetic
energy, then SJ , f
Can you prove this directly without the use of Euler's theorem on homogeneous functions [Problem
11.47]?
LAGRANGE'S EQUATIONS
11.50. (a) Set up the Lagrangian for a one dimensional harmonic oscillator and (6) write Lagrange's
equations. Ans. (a) L = ±mx 2 — ^kx 2 , (b) mx + KX =
11.51. (a) Set up the Lagrangian for a particle of mass m falling freely in a uniform gravitational field
and (b) write Lagrange's equations.
11.52. Work Problem 11.51 in case the gravitational force field varies inversely as the square of the dis
tance from a fixed point O assuming that the particle moves in a straight line through 0.
11.53. Use Lagrange's equations to describe the motion of a particle of mass m down a frictionless in
clined plane of angle a.
11.54. Use Lagrange's equations to describe the motion of a projectile launched with speed v at angle
a with the horizontal.
11.55. Use Lagrange's equations to solve the problem of the (a) twodimensional and (6) three
dimensional harmonic oscillator.
11.56. A particle of mass m is connected to a fixed point P on a horizontal plane by a string of length
I. The plane rotates with constant angular speed w about a vertical axis through a point O of the
plane, where OP = a. (a) Set up the Lagrangian of the system. (6) Write the equations of motion
of the particle.
11.57. The rectangular coordinates (as, y, z) defining the position of a particle of mass tn moving in a force
field having potential V are given in terms of spherical coordinates (r, 6, <f>) by the transformation
equations
x — r sin <f> cos e, y = r sin <£ sin e , z = r cos
Use Lagrange's equations to set up the equations of motion.
Ans.
m[
r — r$ 2 — re 2 cos 2 0] = — —
m
~ d
tz (r 2 0) + r 2 e 2 sin <f> cos <t>
=
1 ay
r Be
m
c
d , „. . „ x 1 d
.. (r 2 e sin 2 </>) = — . . ,
it r sin e a
V
11.58. Work Problem 11.56 if the particle does not necessarily move in a straight line through O.
11.59. Work Problem 4.23, page 102, by use of Lagrange's equations.
LAGRANGE'S EQUATIONS FOR NONHOLONOMIC SYSTEMS
11.60. (a) Work Problem 11.20, page 293, if the paraboloid is replaced by the cone x 2 + y 2 = c 2 z 2 .
(b) What modification must be made to Problem 11.21, page 294, in this case?
11.61. Use the method of Lagrange's equations for nonholonomic systems to solve the problem of a
particle of mass m sliding down a frictionless inclined plane of angle a.
11.62. Work Problem 3.74, page 82 by using the method of Lagrange's equations for nonholonomic
systems.
LAGRANGE'S EQUATIONS WITH IMPULSIVE FORCES
11.63. A uniform rod of length I and mass M is at rest on a horizontal frictionless table. An impulse of
magnitude $ is applied to one end A of the rod and perpendicular to it. Prove that (a) the
velocity given to end A is 4J/M, (b) the velocity of the center of mass is J/M and (c) the rod
rotates about the center of mass with angular velocity of magnitude 6J/MI.
CHAP. Ill
LAGRANGE'S EQUATIONS
307
11.64. In Fig. 1111, AB and BC represent two uniform rods
having the same length I and mass M smoothly hinged
at B and at rest on a horizontal frictionless plane.
An impulse is applied at C normal to BC in the di
rection indicated in Fig. 1111 so that the initial
velocity of point C is v . Find (a) the initial vel
ocities of points A and B and (6) the magnitudes of
the initial angular velocities of AB and BC about
their centers of mass.
Ans. (a) v /7,  2v /7; (6) 3v /ll,  9v /7l
Fig. 1111
11.65. Prove that the total kinetic energy developed by the system of Problem 11.64 after the impulse
is \Mv%
11.66. A square of side a and mass M, formed from 4 uniform rods which are smoothly hinged at their
edges, rests on a horizontal frictionless plane. An impulse is applied at a vertex in a direction
of the diagonal through the vertex so that the vertex is given a velocity of magnitude v . Prove
that the rods move about their centers of mass with angular speed 3v /4a.
11.67. (a) If $ is the magnitude of the impulse applied to the vertex in Problem 11.66, prove that the
kinetic energy developed by the rods is given by 5<5 2 /4ilf. (b) What is this kinetic energy in terms of
i> ? (c) Does the direction of the impulse make any difference? Explain.
11.68. In Problem 11.24, page 296, suppose that the impulse is applied at the center of one of the rods in
a direction which is perpendicular to the rod. Prove that the kinetic energy developed is <jj 2 /8m.
MISCELLANEOUS PROBLEMS
11.69. A particle of mass m moves on the inside of a smooth hollow hemisphere of radius a having its
vertex on a horizontal plane. With what horizontal speed must it be projected so that it will
remain in a horizontal circle at height h above the vertex?
11.70. A particle of mass m is constrained to move inside a
thin hollow frictionless tube [see Fig. 1112] which is
rotating with constant angular velocity a in a hor
izontal xy plane about a fixed vertical axis through
O. Using Lagrange's equations, describe the motion.
11.71. Work Problem 11.70 if the xy plane is vertical.
11.72. A particle of mass m moves in a central force field
having potential V(r) where r is the distance from
the force center. Using spherical coordinates, (a)
set up the Lagrangian and (6) determine the equa
tions of motion. Can you deduce from these equa
tions that the motion takes place in a plane [compare
Problem 5.1, page 121]?
Fig. 1112
11.73. A particle moves on a frictionless horizontal wire of radius a, acted upon by a resisting force which
is proportional to the instantaneous speed. If the particle is given an initial speed v , find the
position of the particle at any time t.
Ans. e = (?nv Ac)(l — e~ Kt/ma ) where e is the angle which a radius drawn to m makes with a fixed
radius such that e = at t = 0, and k is the constant of proportionality.
11.74. Work Problem 11.73 if the resisting force is proportional to the square of the instantaneous speed.
no, \ m J
11.75. A spherical pendulum is fixed at point O but is otherwise free to move in any direction. Write
equations for its motion.
11.76. Work Problem 9.29, page 239, by use of Lagrange's equations.
308
LAGRANGE'S EQUATIONS
[CHAP. 11
11.77. Work Problem 11.20 if the paraboloid of revolution is replaced by the elliptic paraboloid
az = bx 2 + cy 2 where a, b, c are positive constants.
11.78. Prove that the generalized force corresponding to the angle of rotation about an axis physically
represents the component of the torque about this axis.
11.79. (a) Obtain Lagrange's equations corresponding to B and <p in Problem 11.34, page 302, and show
that these are not the same as equations (2) and (3) of that problem. (6) Show how to obtain
equations (2) and (3) of Problem 11.34 from the Lagrange equations of (a).
11.80. Two circular disks, of radius of gyrations K V K 2 and masses m u m 2
respectively, are suspended vertically on a wire of negligible mass [see
Fig. 1113]. They are set into motion by twisting one or both of the
disks in their planes and then releasing. Let X and e 2 be the angles
made with some specified direction.
V//////////////////A
(a) Prove that the kinetic energy is
T
^(m.Kl'e 2 + m 2 K\el)
(b) Prove that the potential energy is
v = iMf + T^*!) 2 ]
where t x and t 2 are torsion constants, i.e. the torques required to
rotate the disks through one radian.
(c) Set up Lagrange's equations for the motion.
Fig. 1113
11.81. Solve the vibrating system of Problem 11.80, finding (a) the normal frequencies and (6) the normal
modes of vibration.
11.82. Generalize the results of Problem 11.80 and 11.81 to 3 or more disks.
11.83. (a) Prove that if m x ¥> m 2 and l x ¥= l 2 in the double pendulum of Problem 11.28, then the normal
frequencies for small oscillations are given by w/2tt where
(m t + m 2 )(h + h) ± V(»h + w 2 ) [«»!(«!  y 2 + m 2 (*i + l 2 ) 2
2l 1 l 2 m 1
(b) Discuss the normal modes corresponding to the frequencies in (a).
9
11.84. Examine the special case h  l 2 , m, ¥* m 2 in Problem 11.83.
11.85. Use Lagrange's equations to describe the motion of a sphere of radius a rolling on the inner surface
of a smooth hollow hemisphere of radius b > a.
11.86. A particle on the inside surface of a frictionless paraboloid of revolution az = x 2 + y 2 at a height
H x above the vertex is given a horizontal velocity v . Find the value of v in order that the particle
oscillate between the planes z = H x and z = H 2 . Ans. v Q = y2gH 2
11.87. Find the period of the oscillation in Problem 11.86.
11.88. A sphere of radius a is given an initial velocity v up a frictionless inclined plane of angle a in a
direction which is not along the line of greatest slope. Prove that its center describes a parabola.
11.89. A bead of mass m is constrained to move on a frictionless horizontal circular wire of radius a
which is rotating at constant angular speed a about a fixed vertical axis passing through a point
on the wire. Prove that relative to the wire the bead oscillates like a simple pendulum.
CHAP. Ill
LAGRANGE'S EQUATIONS
309
11.90. If a particle of mass m and charge e moves with velocity v in an electric field E and magnetic
field B, the force acting on it is given by
F = e(E + v X B)
In terms of a scalar potential * and a vector potential A the fields can be expressed by the relations
E = V*  dA/dt, B = V X A
Prove that the Lagrangian defining the motion of such a particle is
L — «m; 2 + e(A • v) — e*
11.91. Work Problem 10.86, page 278, by use of Lagrange's equations.
11.92. A uniform rod of length I and mass M has its ends constrained to move on the circumference of a
smooth vertical circular wire of radius a > 1/2 which rotates about a vertical diameter with con
stant angular speed u. Obtain equations for the motion of the rod.
11.93. Suppose that the potential V depends on q v as well as q v . Prove that the quantity
T + V
is a constant.
v . dV
dq v
11.94. Use Lagrange's equations to set up and solve the two body problem as discussed in Chapter 5
[see for example page 121.]
11.95. Find the acceleration of the 5 gm mass in the pulley system of
Fig. 1114. Ans. 71^/622
11.96.
A circular cylinder of radius a having radius of gyration K
with respect to its center, moves down an inclined plane of
angle a. If the coefficient of friction is n, use Lagrange's
equations to prove that the cylinder will roll without slipping
K 2
if ii <
tan a. Discuss the cases where n does not
a 2 + K 2
satisfy this inequality.
11.97. Use Lagrange's equations to solve Problem 8.27, page 213.
11.98. Describe the motion of the rods of Problem 11.64 at any time
t after the impulse has been applied.
r~\
15 gm
10 gm
I □ 2gm d
M 7 « m 3gm
5 gm
Fig. 1114
11.99. In Fig. 1115, AB represents a frictionless horizontal plane
having a small opening at O. A string of length I which A_
passes through O has at its ends a particle P of mass m and
a particle Q of equal mass which hangs freely. The particle
P is given an initial velocity of magnitude v at right angles
to string OP when the length OP = a. Let r be the in
stantaneous distance OP while e is the angle between OP and
some fixed line through O.
(a) Set up the Lagrangian of the system.
(6) Write a differential equation for the motion of P in terms of r.
(c) Find the speed of P at any position.
Ans. (a) L = %m[2r 2 + r 2 e 2 ] + mg(l  r)
(6) r = a 2/ i%/r 2 — g
(c) r = y/2av? + 2g(a  r)  2a 2 i%/r
O
IF
Q
Fig. 1115
11.100. Work Problem 11.99 if the masses of particles P and Q are ra x and m 2 respectively.
310
LAGRANGE'S EQUATIONS
[CHAP. 11
11.101. Prove that if v = yfag the particle P of Problem 11.99 remains in stable equilibrium in the circle
r = a and that if it is slightly displaced from this e quilibrium position it oscillates about this
position with simple harmonic motion of period 2irv2ct/3flr.
11.102. Prove that the quantity fy in Problem 11.34, page 302, physically represents the component A 3
of the torque.
11.103. Describe the motion of the system of (a) Problem 11.63 and (6) Problem 11.66 at any time t after
the impulse has been applied.
11.104. Show how to find the angle at which the sphere of Problem 11.37, page 303, falls off.
11.105. (a) Set up the Lagrangian for the triple pendulum of Fig. 1116.
(6) Find the equations of motion.
11.106. Obtain the normal frequencies and normal modes for the triple pendulum of Problem 11.105 assuming
small oscillations.
11.107. Work Problems 11.105 and 11.106 for the case where the masses and lengths are unequal.
m
Fig. 1116
Fig. 1117
11.108. A vertical spring [Fig. 1117] has constant k and mass M. If a mass m is placed on the spring and
set into motion, use Lagrange's eq uations to prove that the system will move with simple harmonic
motion of period 2v^(M + 3w)/3<c.
Chapter 12 HAMILTON IAN
THEORY
HAMILTONIAN METHODS
In Chapter 11 we investigated a formulation of mechanics due to Lagrange. In this
chapter we investigate a formulation due to Hamilton known collectively as Hamiltonian
methods or Hamiltonian theory. Although such theory can be used to solve specific prob
lems in mechanics, it develops that it is more useful in supplying fundamental postulates
in such fields as quantum mechanics, statistical mechanics and celestial mechanics.
THE HAMILTONIAN
Just as the Lagrangian function, or briefly the Lagrangian, is fundamental to Chapter
11, so the Hamiltonian function, or briefly the Hamiltonian, is fundamental to this chapter.
The Hamiltonian, symbolized by H, is defined in terms of the Lagrangian L as
n
H = 2 Vaq a  L (1)
a=l
and must be expressed as a function of the generalized coordinates q a and generalized
momenta p a . To accomplish this the generalized velocities q a must be eliminated from (1)
by using Lagrange's equations [see Problem 12.3, for example]. In such case the function
H can be written
H(pi, .. .,p n , qi, ..., q n , t) (2)
or briefly H(p a , q a , t), and is also called the Hamiltonian of the system.
HAMILTON'S EQUATIONS
In terms of the Hamiltonian, the equations of motion of the system can be written in
the symmetrical form
• dH
dH
qa — t —
dp*
(3)
These are called Hamilton's canonical equations, or briefly Hamilton's equations. The
equations serve to indicate that the p a and q a play similar roles in a general formulation
of mechanical principles.
THE HAMILTONIAN FOR CONSERVATIVE SYSTEMS
If a system is conservative, the Hamiltonian H can be interpreted as the total energy
(kinetic and potential) of the system, i.e.,
H = T + V (4)
Often this provides an easy way for setting up the Hamiltonian of a system.
311
312 HAMILTONIAN THEORY [CHAP. 12
IGNORABLE OR CYCLIC COORDINATES
A coordinate q a which does not appear explicitly in the Lagrangian is called an ignorable
or cyclic coordinate. In such case
*■ = £ = » < 5 >
so that p a is a constant, often called a constant of the motion.
In such case we also have dH/dq a = 0.
PHASE SPACE
The Hamiltonian formulation provides an obvious symmetry between the p a and q a
which we call momentum and position coordinates respectively. It is often useful to imagine
a space of In dimensions in which a representative point is indicated by the In coordinates
(Pu ■ • .,Pn, <7i, . . . , q n ) (6)
Such a space is called a 2n dimensional phase space or a pq phase space.
Whenever we know the state of a mechanical system at time t, i.e. we know all position
and momentum coordinates, then this corresponds to a particular point in phase space.
Conversely, a point in phase space specifies the state of the mechanical system. While the
mechanical system moves in the physical 3 dimensional space, the representative point
describes some path in the phase space in accordance with equations (3).
LIOUVILLE'S THEOREM
Let us consider a very large collection of conservative mechanical systems having the
same Hamiltonian. In such case the Hamiltonian is the total energy and is constant, i.e.,
H(pi, ...,Pn,qi,...,Qn) = constant = E {7)
which can be represented by a surface in phase space.
Let us suppose that the total energies of all
these systems lie between Ei and E 2 . Then the
paths of all these systems in phase space will lie
between the two surfaces H = Ei and H — E 2
as indicated schematically in Fig. 121.
Since the systems have different initial condi
tions, they will move along different paths in the
phase space. Let us imagine that the initial points
are contained in region %i of Fig. 121 and that
after time t these points occupy region % 2 . For
example, the representative point corresponding to
one particular system moves from point A to
point B. From the choice of %i and % 2 it is clear
that the number of representative points in them
are the same. What is not so obvious is the follow Fig. 121
ing theorem called Liouville's theorem.
Theorem 12.1: Liouville's Theorem. The 2n dimensional volumes of %i and ^2 are the
same, or if we define the number of points per unit volume as the density then the density
is constant.
CHAP. 12]
HAMILTONIAN THEORY 313
We can think of the points of %i as particles of an incompressible fluid which move from %i
to %2 in time t.
THE CALCULUS OF VARIATIONS
A problem which often arises in mathematics is that of finding a curve y = Y(x) joining
the points where x — a and x = b such that the integral
>b
f F(x,y,y')dx (8)
*J a
where y'  dy/dx, is a maximum or minimum, also called an extremum or extreme value.
The curve itself is often called an extremal. It can be shown [see Problem 12.6] that a
necessary condition for (8) to have an extremum is
d_/dF\_ dF = (g)
dx\dy'J by v '
which is often called Euler's equation. This and similar problems are considered in a
branch of mathematics called the calculus of variations.
HAMILTON'S PRINCIPLE
The obvious similarity of (9) to Lagrange's equations leads one to consider the problem
of determining the extremals of
J»t 2
L{Qu
, Qn, Ql, . . ., On, t)dt (10)
J»t 2
Ldt
h
where L = T — V is the Lagrangian of a system.
We can show that a necessary condition for an extremal is
dt\dq a J dq a
which are precisely Lagrange's equations. The result led Hamilton to formulate a general
variational principle known as
Hamilton's Principle. A conservative mechanical system moves from time U to time
t 2 in such a way that
C*Ldt (12)
sometimes called the action integral, has an extreme value.
Because the extreme value of (12) is often a minimum, the principle is sometimes referred
to as Hamilton's principle of least action.
The fact that the integral (12) is an extremum is often symbolized by stating that
8 Vhdt = (13)
where S is the variation symbol.
314 HAMILTONIAN THEORY [CHAP. 12
CANONICAL OR CONTACT TRANSFORMATIONS
The ease in solution of many problems in mechanics often hinges on the particular
generalized coordinates used. Consequently it is desirable to examine transformations
from one set of position and momentum coordinates to another. For example if we call
q a and p a the old position and momentum coordinates while Q a and P a are the new position
and momentum coordinates, the transformation is
Pa = Pa (Pi, . . .,Vn, tf 1, . . . , <7n, t), Qa = Qa(Vu • • • , Pn, 01, . . . , q n , t) (14)
denoted briefly by
Pa = Pa (Pa, <?a, t), Qa = Qa(Pa, Qa, t) (15)
We restrict ourselves to transformations called canonical or contact transformations for
which there exists a function Jl called the Hamiltonian in the new coordinates such that
P «Wa> Qa ~^Pa {W)
In such case we often refer to Q a and P a as canonical coordinates.
The Lagrangians in the old and new coordinates are L(p a ,q a ,t) and oC(P a ,Qa,t) re
spectively. They are related to the Hamiltonians H(p a ,q a ,t) and Jl(P a ,Q a ,t) by the
equations
H = 2 Paka ~ L, Ji = 2 PaQa ~ * (17)
where the summations extend from a = 1 to n.
CONDITION THAT A TRANSFORMATION BE CANONICAL
The following theorem is of interest.
Theorem 12.2. The transformation
Pa = Pa(Pa,qa,t), Qa = Qa(Pa,q a ,t) (18)
is canonical if ^p a dq a — ^PadQa (I 9 )
is an exact differential.
GENERATING FUNCTIONS
By Hamilton's principle the canonical transformation (14) or (15) must satisfy the con
ditions that f * L dt and f ' C dt are both extrema, i.e. we must simultaneously have
8 C* Ldt = and sf'^f = (20)
These will be satisfied if there is a function Q such that
4? = L  JZ («)
dt
See Problem 12.11. We call Q a generating function.
By assuming that Q is a function, which we shall denote by d", of the old position co
ordinates q a and the new momentum coordinates P a as well as the time t, i.e.,
CHAP. 12] HAMILTONIAN THEORY 315
Q = effaa Pa, t) (22)
we can prove that [see Problem 12.13]
where P. = M q. = M (w)
Similar results hold if the generating function is a function of other coordinates [see
Problem 12.12].
THE HAMILTONJACOBI EQUATION
If we can find a canonical transformation leading to Jl = 0, then we see from (24) that
P a and Q a will be constants [i.e., P a and Q a will be ignorable coordinates]. Thus by means of
the transformation we are able to find p a and q a and thereby determine the motion of the
system. The procedure hinges on finding the right generating function. From the third
equation of (23) we see by putting Jl = that this generating function must satisfy the
partial differential equation
^ + H(p a , q a , t) = (25)
f + *(f. «.«) = ° <«•>
This is called the HamiltonJacobi equation.
SOLUTION OF THE HAMILTONJACOBI EQUATION
To accomplish our aims we need to find a suitable solution of the HamiltonJacobi
equation. Now since this equation contains a total of n + 1 independent variables, i.e.
Qu qz, . . . , q n and t, one such solution called the complete solution, will involve n + 1 con
stants. Omitting an arbitrary additive constant and denoting the remaining n constants by
Pu (32, . . . , p n [none of which is additive] this solution can be written
of = offai, tf2, ... , q n , pu fa, ..., £„, *) (27)
When this solution is obtained we can then determine the old momentum coordinates by
def
Also, if we identify the new momentum coordinates P a with the constants p a , then
M
dfia
O — doJ* _
where y a , a = 1, . . . , n are constants.
Using these we can then find q a as functions of p a , y a and t, which gives the motion of
the system.
CASE WHERE HAMILTONIAN IS INDEPENDENT OF TIME
In obtaining the complete solution of the HamiltonJacobi equation, it is often useful to
assume a solution of the form
316 HAMILTONIAN THEORY [CHAP. 12
of = Si(qi) + S 2 (g 2 ) + • • • + Sn{q n ) + F(t) (30)
where each function on the right depends on only one variable [see Problems 12.15 and
12.16]. This method, often called the method of separation of variables, is especially useful
when the Hamiltonian does not depend explicitly on time. We then find that F(t) = —Et,
and if the time independent part of of is denoted by
5 = Si(qi) + S2M + ••■ + S n {q n ) (31)
the HamiltonJacobi equation (26) reduces to
*(£•«■) = E (S2)
where E is a constant representing the total energy of the system.
The equation (32) can also be obtained directly by assuming a generating function S
which is independent of time. In such case equations (23) and (24) are replaced by
Pa = S , Q a = j§, JH = H = E (33)
dq a OJTa
where Pa = ~tq~> Q<* — Tp~ (**)
PHASE INTEGRALS. ACTION AND ANGLE VARIABLES
Hamiltonian methods are useful in the investigation of mechanical systems which are
periodic. In such case the projections of the motion of the representative point in phase
space on any p a q a plane will be closed curves C«. The line integral
J a = <f> Padq a ( 35 )
J c a
is called a phase integral or action variable.
We can show [see Problems 12.17 and 12.18] that
S = S(qi, ...,qn,Ju ...,Jn) (*6)
where p. = Jg, «« = J ^
It is customary to denote the new coordinates Q a by w a so that equations (37) are re
pIacedby as bs (m
Thus Hamilton's equations become [see equations (33) and (34)]
dW a dJ a
where Jl = E in this case depends only on the constants J a . Then from the second equa
tion in (39), nn s
V } Wa = fat + Ca (*°)
where /« and c a are constants. We call w a angle variables. The frequencies /« are given by
dJa
See Problems 12.19 and 12.20.
CHAP. 12]
HAMILTONIAN THEORY 317
Solved Problems
THE HAMILTONIAN AND HAMILTON'S EQUATIONS
12.1. If the Hamiltonian H = 2 V«qa  L, where the summation extends from
a = 1 to n, is expressed as a function of the coordinates q a and momenta p a , prove
Hamilton's equations, ^ vrr ^ d jj
V" = 'Wee' *" = W*
regardless of whether H (a) does not or (b) does contain the variable time t explicitly.
(a) H does not contain t explicitly.
Taking the differential of H  2 Va Qa — L > we have
dH = 2P a <*<7« + 2?«# a  ^^dq a  1jrdi a (1)
Then using the fact that p a = dL/dq a and p a = dL/dq a , this reduces to
dH  *2,q a dp a  2p a dq a {2)
But since H is expressed as a function of p a and q a , we have
dH = 2jrdp a + 1^dq a (3)
dp a dq a
Comparing (2) and (3) we have, as required,
. _ dH_ . BH
q <*  dp a > Pa dq a
(b) H does contain t explicitly.
In this case equations (1), (2) and (3) of part (a) are replaced by the equations
dH = 2?>a<4* + 2g«<*P«  2f^<*<7«  2^<*<L  ~^dt U)
dH = 2q a dp a  *2>p a dq a  — dt (5)
m = 2 £ dp „ + 2 f^ + ff ,* «
Then comparing (5) and (6), we have
. _ dH_ . _ _d# 8H _ _3L
9 « ~ dp a ' Pa ~ dq tt ' dt dt
12.2. If the Hamiltonian H is independent of t explicitly, prove that it is (a) a constant and
is (b) equal to the total energy of the system.
(a) From equation (2) of Problem 12.1 we have
jfe = 2 ka Pa ~ 2 Pa Qa =
Thus H is a constant, say E.
(6) By Euler's theorem on homogeneous functions [see Problem 11.47, page 305],
dq a
318
HAMILTONIAN THEORY
[CHAP. 12
where T is the kinetic energy. Then since p a  dL/dq a = dT/dq a [assuming the potential V
does not depend on q a ], we have 2p a <Za = 2T. Thus as required,
H = 2,PaQa~L = 2T(TV) = T + V = E
12.3. A particle moves in the xy plane under the influence of a central force depending only
on its distance from the origin, (a) Set up the Hamiltonian for the system. (6) Write
Hamilton's equations of motion.
(a) Assume that the particle is located by its polar coordinates (r, e) and that the potential due to
the central force is V(r). Since the kinetic energy of the particle is T = £m(r 2 + r 2 * 2 ), the
Lagrangian is .„„•„,,,, „
L = T  V = im(r 2 + rW)  V(r)
We have
so that
Pr
dL/dr = mr, p g = dL/de — mr 2 6
r = p r /m, o = p e /wr 2
Then the Hamiltonian is given by
H = £ p a q«  L = P r r + PeO  {$m(r* + r&)  V(r)>
= Pr S +" 3  4*
Pe
P!. + r2 .^LV V(r)
m 2 m 2 W * '
2m 2mr 2 '
Note that this is the total energy expressed in terms of coordinates and momenta.
(6) Hamilton's equations are q a = dH/dp a , p a = dH/dq a
Thus ; = dH/dp r = p r lm, h = dH/dpe  p Q /mr*
p r = dH/dr = ppmr*  V(r), p e = dH/de =
Note that the equations (5) are equivalent to the corresponding equations (3).
(1)
(2)
(3)
(4)
(5)
(6)
PHASE SPACE AND LIOUVILLE'S THEOREM
12.4. Prove Liouville's theorem for the case of one degree of freedom.
We can think of the mechanical system as
being described in terms of the motion of rep
resentative points through an element of vol
ume in phase space. In the case of a mechanical
system with one degree of freedom, we have a
two dimensional (p, q) phase space and the vol
ume element reduces to an area element dpdq
[Fig. 122].
Let p = p(p, q, t) be the density of rep
resentative points, i.e. the number of repre
sentative points per unit area as obtained by
an appropriate limiting procedure. Since the
speed with which representative points enter
through AB is q, the number of representative
points which enter through AB per unit time is
pq dp (1)
The number of representative points which leave through CD is
B(q, p + dp) C(q + dq, P + dp)
#■ *
.. I »»
Mq,v)
Fig. 122
P q + j (pq) dq f dp
D(q + dq, p)
CHAP. 12] HAMILTONIAN THEORY 319
Thus the number which remain in the element is (i) minus (2), or
—^(pq)dpdq (3)
Similarly the number of representative points which enter through AD and leave through BC are
respectively , .
pp dq and <pp + — (pp) dp > dq
Thus the number which remain in the element is
 — (pp)dpdq (4)
The increase in representative points is thus [adding (3) and (4)]
{*#+^V**
Since this is equal to ^dpdq, we must have
dp + [ Kpq) + d(pfr )
dt 1 dq dp
dp , dq , dp • , dp . op . .
tt + PF"+ fq + P^ + rp = (5)
dt dq dq * dp dp v
Now by Hamilton's equations p = —dH/dq, q = dH/dp so that
dp _ dm dq _ dm
dp dpdq' dq dq dp
Thus since we suppose that the Hamiltonian has continuous second order derivatives, it follows
that dp/dp = —dq/dq. Using this in (5), it becomes
But this can be written dp/dt = (7)
which shows that the density in phase space is constant and thus proves Liouville's theorem.
12.5. Prove Liouville's theorem in the general case.
In the general case the element of volume in phase space is
dV = dq y • • • dq n dp t • • • dp n
In exactly the same manner as in Problem 12.4 the increase of representative points in dV is found
to be
(3(p«i) , , d(pq n ) a(pp t ) d(pP n )\ _.
— < — (•••+ — 1 — r 1 • • • H > dV
{ dq t dq n 3p! dp n J
and since this is equal to £dV, we must have
dt
dp , d( pqi ) d(pq n ) d( pPl ) d(pp n )
— f — _ \. . . . __ — J _^ j. . . . __ — _ y
dt dq x dq n dp t dp n
dp , £ a <Pg«) . x d ^)
dt a =l dq a «=i dp a
This can be written as
dp , £ (dp . , dp • \ , « fdq a ,dp a \
320 HAMILTONIAN THEORY [CHAP. 12
Now by Hamilton's equations p a = —dH/dq a , q a = dH/dp a so that
dp a _ d 2 H dq a _ d 2 H
^K ~~ ~ d P<x d<la ' dPa ~ d <I a d Pa
Hence dp a /dp a = dq a /dq a and (1) becomes
&+ 2 (£«« + s^O = o («)
at a=l \^a "Pa /
i.e., <V<tt = (#)
or p = constant.
Note that we have used the fact that if p = p(q u . . . , q n > Pi, ■ ■ ■ > Pn> *) then
dp 4j / dp dq a dp dp a \ dp _ » fdp_. ,dp_ \ , dp
CALCULUS OF VARIATIONS AND HAMILTON'S PRINCIPLE
F(a, y, y') dx to be an extremum [maxi
mum or minimum] is
d L (dF_\_dF = Q
dx \ by' ) dy
Suppose that the curve which makes I an extremum is given by
y = Y(x), a^x^b CO
Then V = Y(x) + e V (x) = Y + e V <*)
where e is independent of x, is a neighboring curve through x = a and x = b if we choose
via) = v(b) = W
The value of 7 for this neighboring curve is
F(x, Y + ev , Y' + e V ') dx U)
a
jr I
This is an extremum for e = 0. A necessary condition that this be so is that ^  e=o = 0. But by
differentiation under the integral sign, assuming this is valid, we find
de  £ =o J a \dy v ^ dy' v J
which can be written on integrating by parts as
 £'{£=(5)}*  °
where we have used (3). Since v is arbitrary, we must have
dF _d_/dF\ _ n nr ±{§L\.W = o
~dy dx\dy'J ~ u dx\dy'J dy
which is called tf wtor** or Lagrange's equation. The result is easily extended to the integral
Xb
F{X, i/i, If J, 2/2» 1/2. • • ■ » Vn» »n) d *
u
and leads to the Euler's or Lagrange's equations
CHAP. 12]
HAMILTONIAN THEORY
321
dx\by' a J dy a
a = 1,2, ...,n
By using a Taylor series expansion we find from (4) that
/( e ) _ j(o) = e j f j v + —T)')dx + higher order terms in e 2 , e 3 , etc.
The coefficient of e in (5) is often called the variation of the integral and is denoted by
F(x, y, y') dx
(5)
■f.
F(x, y, y') dx is an extremum is thus indicated by
a b
8 f F(x, y, y') dx =
12.7. Discuss the relationship of Hamilton's principle with Problem 12.6.
By identifying the function F{x,y,y') with the Lagrangian L(t,q,q) where x,y and y' are re
placed by t, q, q respectively, we see that a necessary condition for the action integral
r
Ldt
(1)
to be an extremum [maximum or minimum] is given by
l(f)f = ° < 2)
Since we have already seen that (2) describes the motion of a particle, it follows that such motion
can also be achieved by requiring that (1) be an extremum, which is Hamilton's principle.
For systems involving n degrees of freedom we consider the integral (1) where
L = L(t, q v q lt q 2 , q 2 , . . . , q n , q n )
which lead to the Lagrange equations
dt\dq a J dq a
=
a = 1,2, . . .,«
A(« »Vo)
12.8. A particle slides from rest at one point on a
f rictionless wire in a vertical plane to another
point under the influence of gravity. Find the
total time taken.
Let the shape of the wire be indicated by curve
C in Fig. 123 and suppose that the starting and fin
ishing points are taken to be the origin and the point
A{xQ,yo) respectively.
Let P(x, y) denote any position of the particle
which we assume has mass m. From the principle
of conservation of energy, if we choose the horizontal
line through A as reference level, we have
Potential energy at O + kinetic energy at O = potential energy at P + kinetic energy at P
or mgy + = mg(y — y) + %m(ds/dt) 2
where ds/dt is the instantaneous speed of the particle at time t. Then
ds/dt = ±Vzgy (0
If we measure the arc length s from the origin, then s increases as the particle moves. Thus ds/dt
is positive, so that ds/dt = y/2gy or dt = ds/y/2gy.
The total time taken to go from y = to y = y is
Fig. 123
322 HAMILTONIAN THEORY [CHAP. 12
T = I dt = I
^0 ^=0
'2gy
But (cte) 2 = (do;) 2 + (dy) 2 or cte = ^1 + y> 2 dx. Thus the required time is
1 r v Vi + y' 2
r = —— I j= dx (2)
V2gJ y=0 vv
12.9. If the particle of Problem 12.8 is to travel from point O to point A in the least pos
sible time, show that the differential equation of the curve C denning the shape of
the wire is 1 + y' 2 + lyy" = 0.
A necessary condition for the time r given by equation (2) of Problem 12.8 to be a minimum
is that
±(d_F_\_8F =
dx\dy'J dy U K1)
where F = (1 + y' 2 ) 1 ' 2 V V2 (2)
Now BFlby' = (l + y' 2 ) 1/2 y'y^ 2 , dF/dy = — 1(1 + y' 2 )^ y s/2
Substituting these in (1), performing the indicated differentiation with respect to x and simplifying,
we obtain the required differential equation.
The problem of finding the shape of the wire is often called the brachistochrone problem.
12.10. (a) Solve the differential equation in Problem 12.9 and thus (b) show that the required
curve is a cycloid.
(a) Since x is missing in the differential equation, let y' = u so that
„ _ du _ dw dy_ _ du , _ du
dx dy dx dy dy
Then the differential equation becomes
l + u* + 2yu ( ^=0 or ^ + ^ =
dy 1 + u z y
Integration yields
In (1 + u 2 ) + In y = In 6 or (1 + u 2 )y = b
where 6 is a constant. Thus
 = * = t  J ^
\ y
since the slope must be positive. Separating the variables and integrating, we find
Letting y — b sin 2 e, this can be written
x = f J b sin2 9 . 26 sin ecosede + c
J V 6 cos 2 e
= 26 f sin 2 6 de + c = 6 j (1  cos 2e) de + c = £6(2*  sin 2e) + c
Thus the parametric equations of the required curve are
x = £6(2*  sin 2d) + c, y = b sin 2 e = 6(1  cos 2e)
Since the curve must pass through the point x = 0, y = 0, we have c = 0. Then letting
= 2(9, a = 6 CO
the required parametric equations are
x = a(<p — sin <f>), y = a(l — cos <p) (2)
CHAP. 121
HAMILTONIAN THEORY
323
(b) The equations (2) are parametric equations of a cycloid [see Fig. 124]. The constant a must
be determined so that the curve passes through point A. The cycloid is the path taken by a
fixed point on a circle as it rolls along a given line [see Problem 12.89].
7<
/ \
CANONICAL TRANSFORMATIONS AND GENERATING FUNCTIONS
12.11. Prove that a transformation is canonical if there exists a function q such that
dq/dt = LJi.
The integrals I Ldt and I £dt must simultaneously be extrema so that their varia
tions are zero, i.e.,
Thus by subtraction,
t.
! \ Ldt  and SJ C dt =
f
(L  £) dt
This can be accomplished if there exists a function Q such that
L  C = d£/ctt
rdt = S{£(£ 2 ) — £(*i)} = o
The function Q is called a generating function.
12.12. Suppose that the generating function is a function T of the old and new position co
ordinates q a and Q a respectively as well as the time t, i.e. T = T(q a , Q a , t). Prove that
Va = , P a = *[1, ^ = ^ + tf where P« =
dQ a '
dt
dQa ' dPa
By Problem 12.11,
dT
dt
= Lj£ = 2, Pa q a  H  V2P a Q a  Si
= Sp«g« ~ 2P«Q a + Si  H
dT = 2p a dq a  2P a dQ a + (Si~H)dt
But if T = T{q a , Q a , t), then
39o
Comparing (1) and (2), we have as required
dT
dT = 2£*Ia+ 2^dQ a + £dt
3Q«
dt
Pa 
dq a '
P„ = 
3t
*c#
 _ djj
Qa ~ 8P„
(1)
The equations * « — aQ '
follow from the fact that Si is the Hamiltonian in the coordinates P a , Q a so that Hamilton's equa
tions hold as in Problem 12.1.
324 HAMILTONIAN THEORY [CHAP. 12
12.13. Let efbea generating function dependent only on q a , P a , t. Prove that
Va ~W«' Qa = eK' J( = lt +H where Pa = Wa' Qa = W~ a
From Problem 12.12, equation (1), we have
dT = ^p a dq a  ^,P a dQ a + (J(H)dt
Or ^T + ^PaQa) = 2Pa^a + 2 Q« <*P« + (Jl~H)dt (1)
i.e., def = 2p«dg a + 2Q«dP a + (J(H)dt (2)
where gf = T + ^P a Q a (3)
But since gf is a function of q a , P a , t,
^ = 2^^«+ SJ^p. + J^ «)
Comparing (2) and (4), 3 , fl f a ,
The results P a = 777, Q a = 75
follow as in Problem 12.12, since J# is the Hamiltonian.
12.14. Prove that the transformation P = i(p 2 + q 2 ), Q = tan 1 (q/p) is canonical.
Method 1.
Let the Hamiltonians in the coordinates p, q and P, Q be respectively H(p, q) and J#(P, Q) so
that H(p, q) — Jfl(P, Q). Since p, q are canonical coordinates,
• dH • 3jF/ /^x
P = —~r , Q  7— (■*)
dq dp
; = i^H4, i = *> + *« »
dH ^ 9J(dP dj£dQ dH = ?MdP 3JtdQ ()
dq dP dq dQ dq' dp dP dp dQ dp
From the given transformation equations we have
5P _ dP __ dQ __ q dQ _ p
dp ~ V ' dq ~ q ' dp p 2 + q 2 ' dq p 2 + q 2
Also, differentiating the transformation equations with respect to P and Q respectively, we find
* = »§ + <§• • = (>&<&)A + «*>
Solving simultaneously, we find
dp __ P dq.  9. *L   a M. = v U)
dP " p 2 + q 2 ' dP ~ p 2 + q 2 ' dQ q ' dQ V
CHAP. 12] HAMILTONIAN THEORY 325
Then equations (1) and (2) become
sh _ djj p djj en BJj q Bjj
dq ~ Q dP P 2 + q 2 dQ ' dp P dP p 2 + q 2 dQ K '
Thus from equations (1), (5) and (6) we have
P
qQ
sji p ajl
p 2 + q 2* ™ H 8P p 2 + q 2 dQ
q JL . A 9Jl Q dj(
; P + pQ = p
p 2 + g 2* r ** ~ * d p p 2 + q2 dQ
Solving these simultaneously we find
. _ _ej( . _ 3_M
F " dQ ' * ~ dP
which show that P and Q are canonical and that the transformation is therefore canonical.
(7)
Method 2.
By Theorem 12.2, page 314, the transformation is canonical if
1p a dq a  J,P a dQ a (8)
is an exact differential. In this case (8) becomes
pdq  PdQ = pdq  ±(p 2 + q 2 ) ( Pdqqdp
\ P + 9
= %(P dq + q dp) = d{±pq)
an exact differential. Thus the transformation is canonical.
THE HAMILTONJACOBI EQUATION
12.15. (a) Write the Hamiltonian for the one dimensional harmonic oscillator of mass m.
(b) Write the corresponding HamiltonJacobi equation, (c) Use the HamiltonJacobi
method to obtain the motion of the oscillator.
(a) Method 1.
Let q be the position coordinate of the harmonic oscillator, so that q is its velocity. Since
the kinetic energy is T = ^mq 2 and the potential energy is V = ^*cg 2 , the Lagrangian is
L = T  V = \mq 2  \ K q 2 (1)
The momentum is p = dL/dq = mq (2)
so that q = p/m (3)
Then the Hamiltonian is
H = *2,p a q a  L = pq  (%mq 2  % K q 2 )
= £p 2 /m + % K q 2 U)
Method 2.
By Problem 12.2, since the Hamiltonian is the same as the total energy for conservative
systems,
H = ±mq 2 + \nq 2 — \m{plm) 2 + £«<7 2 = \p 2 lm + ±ieq 2
326 HAMILTONIAN THEORY [CHAP. 12
(b) Using p = dof/dq and the Hamiltonian of part (a), the HamiltonJacobi equation is [see equa
tion (26), page 315]
(c) Assume a solution to (5) of the form
ef = SM + S 2 (t) (6)
1 /dS x \ 2 dS 2
Then (5) becomes 2^W7 + ^ = ~~dt (7)
Setting each side equal to the constant /?, we find
whose solutions, omitting constants of integration, are
Si = f V2m( i 8  l/cg 2 ) ^ s 2 = pt
so that (6) becomes ^ = j V2m(/3  ^/cg 2 ) dq ^ pt
Let us identify /? with the new momentum coordinate P. Then we have for the new position
coordinate,
(8)
(9)
_ B<£ _ _§_
Q ~ dp dp
\ f y/2m(p  ^/cg 2 ) dq  0*
V2m r dq
C dq
J v/rt —
2 J y/p  %Kq*
But since the new coordinate Q is a constant y,
ViJm /" dq
 t
s
y/p  &Q 2
 t = y
or on integrating, VW* sin" 1 (qy/ic/2p )  t + y
Then solving for g, q = y/2plic sin yJTJm (t + y) (* )
which is the required solution. The constants p and y can be found from the initial conditions.
It is of interest to note that the quantity p is physically equal to the total energy E of the
system [see Problem 12.92(a)]. The result (9) with p = E illustrates equation (SI) on page 316.
12.16. Use HamiltonJacobi methods to solve Kepler's problem for a particle in an inverse
square central force field.
The Hamiltonian is H = ^ \ p ? + ~^ J ~ ~r
Then since p r = d^/dr, p e = d^/de, the HamiltonJacobi equation is
It + 2m \V Br J ^ r 2 V Be J J
Let J = S 1 (r) + S 2 (e) + S 3 (t) W
* =
CHAP. 12] HAMILTONIAN THEORY 327
1 f/dSA 2 , 1 f dS A\ K _ dS s
Then (2) becomes ^ ^r J + ^{^fj J " ~ ~ "^
Setting both sides equal to the constant /? 3 , we find
dS z ldt = /? 3 (4)
^{(^) 2 ^(f )}  f  *
Integration of (4) yields, apart from a constant of integration,
S 3 = /? 3 *
Multiply both sides of (5) by 2mr 2 and write it in the form
dv) = r2 \ 2m ^ + r{dir
Then since one side depends only on e while the other side depends only on r, it follows that each
side is a constant. Thus
dS 2 /de = p 2 or S 2 = P 2 (0)
r \ dr
dS t
(8)
(9)
ui j = V2m/3 3 + 2mK/r  0§/r* (7)
on taking the positive square root. Then
S t = f \/2m/?3 + 2raK/r  /3^/r 2 dr
Thus gT = J V2m/? 3 + 2mtf/r  0*/r 2 dr + /3 2 e  (3 3 t
Identifying /? 2 and /? 3 with the new momenta P r and P e respectively, we have
Qr = Wz = W~2 S ^ 2m ^3 + 2mK/r  /? 2 /r 2 dr + e = Yi
Qe = §7 = af~ J* ^ 2m/?3 + 2mX/r ~ % /r2 dr ~ * = 72
since Q r and <? fl are constants, say y 1 and y 2 . On performing the differentiations with respect to
/3 2 and /? 3 , we find
«/ 7"'
Pidr
• 2 \/2m^3 + 2mX/r  jSf/r 2
»  7i (JO)
, „ = = * + 7 2 (**)
V2m/? 3 + 2mK/r  /3 2 /r 2
The integral in (10) can be evaluated by using the substitution r = II u, and after integrating we
find as the equation of the orbit,
r = 2 = («)
1  Vl + ZPzpymK* cos (* + B/2  y x )
The constant /? 3 can be identified with the energy E [see Problem 12.92(b)], thus illustrating equation
{31), page 316. If E = /3 3 < 0, the orbit is an ellipse; if E = 3 = 0, it is a parabola; and if
E — /3 3 > 0, it is a hyperbola. This agrees with the results of Chapter 5.
The equation (11) when integrated yields the position as a function of time.
328 HAMILTONIAN THEORY [CHAP. 12
PHASE INTEGRALS AND ANGLE VARIABLES
12.17. Let gT be a complete solution of the HamiltonJacobi equation containing the n constants
(3i, . . . , /?„. Let J a = y p a dq a . Prove that the J a are functions of the /? a only.
We have ^ = S 1 (q l , lt . . ., n ) + ••• + S n (q n , fi lt . . . , (3 n )  p x t (1)
where the constant p 1 —E, the total energy. Now
3<2f dS a
dq a dq a v ;
Thus J a = § Padq a = §~dq a {S )
But in this integration q a is integrated out, so that the only quantities remaining are the con
stants Pi, . . ., p n . Thus we have the n equations
J a = J a (fiv • ■ ■ > Pn) a = l,...,n (4)
Using (4) we can solve for p u , ..,p n in terms of J u .. .,J n and express (1) in terms of the J a .
12.18. (a) Suppose that the new position and momentum coordinates are taken to be w a and
J a respectively. Prove that if Ji is the new Hamiltonian,
J a = dM/dWa, W a = dJl/dJ a
(b) Deduce from (a) that
Ja = constant and w a — fat + c a
where f a and c a are constants and f a = dJl/dJ a .
(a) By Hamilton's equations for the canonical coordinates Q a ,P„,
Pa = dJ(/dQa, Qa = dJ{/dP a (1)
Then since the new position and momentum coordinates are taken as Q a — w a and P a = J a >
these equations become
J a = dJl/dWa, W a = dJl/dJ a (2)
(b) Since ^[ — E, the new Hamiltonian depends only on the J a and not on the w a . Thus from (2)
we have
J a = 0, w a = constant = f a (3)
where f a = djK/dJ a  From (3) we find, as required,
J a = constant, w a — f a t + c a (4)
The quantities J a are called action variables while the corresponding integrals
J Padq a = J a (5)
where the integration is performed over a complete cycle of the coordinate q a , are called
phase integrals. The quantities w a are called angle variables.
12.19. (a) Let Aw a denote the change in w a corresponding to a complete cycle in the particu
lar coordinate q r . Prove that
1 if a = r
Aw a = i
if a ¥^r
CHAP. 12] HAMILTONIAN THEORY 329
(b) Give a physical interpretation to the result in (a).
9 C dS , dJ r fl if a = r
' M a yw r dqr = — =
(a) Aw a
5 J a [ if a # r
where we have used the fact that w a = aS/d«7 a [see Problems 12.17 and 12.18] and have as
sumed that the order of differentiation and integration is immaterial.
(6) From (a) it follows that w a changes by one when q a goes through a complete cycle but that
there is no change when any other q goes through a complete cycle. It follows that q a is a
periodic function of w a of period one. Physically this means that the f a in equation (4) of
Problem 12.18 are frequencies.
12.20. Determine the frequency of the harmonic oscillator of Problem 12.15.
A comple te cy cle of the c oordin ate q [see equation (10), Problem 12.15] consists in the motion
from q = y/2(3/ K to q = +V2/3//C and back to q = y/2ph. Then the action variable is
= J pdq = 2 J y/2m(/3  \ K q*) dq = 4 J y/2m(p  % K q*) dq
V2/3/K
' = * = £>£ = .* «* ' = °4 = i{i
= 27T/?VW/C
Thus
12.21. Determine the frequency of the Kepler problem [see Problem 12.16].
^ A complete cycle of the coordinate r consists in the motion from r = r min to r max and back to
r ~ r min> where r min and r max are the minimum and maximum values of r given by the zeros
of the quadratic equation [see equation (10), Problem 12.16]
2m/? 3 + 2mK/r  ppr 2 = (1)
We then have from equations (6) and (7) of Problem 12.16,
J 6 = § Ve de = § d ^ds = §^de = f* fi%d , = u Pt (f)
r r d<of rds< r rmax .
J r = J> Pr dr = y^rdr = Jjdr = 2 J V2m/? 3 + 2mK/r  &/r* dr
r min
= 2wmK/y/2m/3 3  2tt/3 2 ^
From (2) and (3) we have on elimination of p 2 ,
J e + J r = 2TrimKly]— 2m/3 3 m
Since p 3 = E, (4) yields
_ 2^2 m g2 2^2 m /f2
^"OVfT^ othat jr = __
Then the frequencies are
J6 dJ 9 (./«, + J r )3' fr dJ r ~ (J e +«7 r )3
Since these two frequencies are the same, i.e. there is only one frequency, we say that the
system is degenerate.
330 HAMILTONIAN THEORY [CHAP. 12
MISCELLANEOUS PROBLEMS
12.22. A particle of mass m moves in a force field of potential V. Write (a) the Hamiltonian
and (b) Hamilton's equations in spherical coordinates (r, 9, </>).
(a) The kinetic energy in spherical coordinates is
T = \m(r 2 + r 2 e 2 + r 2 sin 2 9 2 ) (1)
Then the Lagrangian is
L = T V = \m(r 2 + r 2 e 2 + r 2 s i n 2 tf £2) _ y( r , ^ ^ ^
We have
p r — dL/dr = mf , p e = dL/do = mr 2 8, p^, = 3L/30 — mr 2 sin 2 (5)
, Pr • Pe • P<*>
and r = — , = — „, = — 2 . 2 „ (4)
m mr mr sin 2 v '
The Hamiltonian is given by
H = ^tP a q a — L
= p r r + p e e + pqQ — ^m(r 2 + r 2 e 2 + r 2 sin 2 e 2 ) + V(r, e, 0)
n 2 m 2 « 2
_ Pr_ , Pe , P<& v , ,
2m 2mr 2 2mr 2 sin 2 <? + v &>'>*) i 5 )
where we have used the results of equations (4).
We can also obtain (5) directly by using the fact that for conservative systems the
Hamiltonian is the total energy, i.e. H = T + V.
(b) Hamilton's equations are q a —  — , p a = — t — . Then from part (a),
dp a dq a
dH Pe . dH P4
. _ dH_ _ Pr • _ dH_ _ _Po_ . _ dH_ _
dp r m ' dp e mr 2 ' dp$ mr 2 sin 2
* = _^ = EL + p * _ §Z
r dr mr 3 mr 3 sin 2 e dr
. _ _dH_ _ P% cos dV
P° ~ do ~ mr 2 sin 3 e do
' = ^R  _§¥_
P * ~~ 30 30
12.23. A particle of mass m moves in a force field whose potential in spherical coordinates
is V — — (K cos 0)/r 2 . Write the Hamilton Jacobi equation describing its motion.
By Problem 12.22 the Hamiltonian is
N  1 ( a * I P ° I P * ) Kc0Sff (1)
n ~ 2m V r r 2 r 2 sin 2 0/ r 2 K '
AqJ" 3<of 9(of
Writing p r — r— , p e = r— , p^ — — — , the required Hamilton Jacobi equation is
Ml. 1 iYMY. ^^Vi 1 /^Y) gcosg =
3* + 2m U 3r / r 2 V 3* / r 2 sin 2 \ 30 J r
12.24. (a) Find a complete solution of the Hamilton Jacobi equation of Problem 12.23 and
(b) indicate how the motion of the particle can be determined.
(a) Letting ©f = S x (r) + S 2 (&) + £3(0) ~ Et in equation (2) of Problem 12.23, it can be written
CHAP. 12] HAMILTONIAN THEORY 331
(1)
i fds t y , i /ds 2 y i /ds 3 y Kcos9
2m\dr J r 2mr 2 \ do J 2rar 2 sin 2 e\d<f>) r 2 ~
Multiplying equation (1) by 2mr 2 and rearranging terms,
/rfSiV /dS 2 \ 2 1 /dSo\2
Since the left side depends only on r while the right side depends on e and <f>, it follows that
each side must be a constant which we shall call /?j. Thus
\d7) ~ 2mEr2 = Pl (2)
/dS 2 \ 2 1 /dSo\2
and (^j a^^j +fcOr««. = fc W
Multiplying equation (5) by sin 2 * and rearranging terms,
/dS 3 y /dS a \*
\~d&) = 2mKain20cos9  /^sin 2 * — sin 2 * ( —j— J (4)
Since the left side depends only on <p while the right side depends only on e each side must be a
constant which we can call /3 2 . However, since
^ = 1? = ~dj < 5 >
we can write /3 2 = p^. This is a consequence of the fact that is a cyclic or ignorable co
ordinate. Then (4) becomes
2mK sin 2 e cos 6 — p x sin 2 e  sin 2 e ( ~ ) = p (6)
By solving equations (2), (6) and (5), we obtain
s i  J y/2mE + pjr 2 dr, S 2 = J yJImK cos  p 2 esc 2 *  /?i ete, S 3 = p^
where we have chosen the positive square roots and omitted arbitrary additive constants. The
complete solution is
of = J V2mE~+JJr^dr + J v / 2m J K' cos *  p$ esc 2 e  /3 1 de + p^  #<
(6) The required equations of motion are found by writing
30J ~ Yl ' ~dE ~ l2 ' dp^ = Y3
and then solving these to obtain the coordinates r, 0,0 as functions of time using initial condi
tions to evaluate the arbitrary constants.
12.25. If the functions F and G depend on the position coordinates q a , momenta p a and time
t, the Poisson bracket of F and G is denned as
\f G] = y (**L!*L  dF dG )
o \dp*dq a dq a dpaj
Prove that (a) [F,G] = [G,F], (6) [Fi + F 2 ,G] = [F U G] + [F*,G], (c) [F,q r ] =
dF/dpr, (d) [F,p r ] = dF/dq r .
(a) \F,G\ = ^(**L*G*LdG\ _ _^(dG_dL_dGdF_\ _ _ {rp]
a \3p« 3g a 9g a ap^/ a \dp a 3g a dq a dp a J
This shows that the Poisson bracket does not obey the commutative law of algebra.
332 HAMILTONIAN THEORY [CHAP. 12
(h) ]F+F G] _ y P>(* T i + * , 2) 8G a{F l + FJ dG \
(b) [F 1 + F 2 ,G]  i dpa dga dga dp ^
V ( d ll*<L _ M\ 8G_\  /^2 ^G _ Mj, _3G
= [^i,G] + [F 2 ,G]
This shows that the Poisson bracket obeys the distributive law of algebra.
^ f dF dq r dF dq r \ _ qF_
{C) [,<lrl ~ %\d Pa dq a dq a dp a J d Pr
since dq r /dq a = 1 for a = r and for a ¥= r, while dq r /dp a = for all a. Since r is ar
bitrary, the required result follows.
(d)
[ ,Pr] ~ a\dp a dq a dq a dp a J dq r
since dp r /dq a = for all a, while dp r /dp a = 1 for « = r and for a ¥> r. Since r is ar
bitrary, the required result follows.
12.26. If H is the Hamiltonian, prove that if / is any function depending on position, momenta
and time, then ,, «*
or dt ~ at + r \dq a q « + dp a Pa
dH • _ dH (*\
But by Hamilton's equations, q a — T~r > Pa — ~TZ~ K '
dp a dq a
Then (2) can be written
§L = U. + 2 f^ ^L& —) = &+ [H, f]
dt dt H \dq a dp a dp a dqj dt
Supplementary Problems
THE HAMILTONIAN AND HAMILTON'S EQUATIONS
12.27. A particle of mass m moves in a force field of potential V. (a) Write the Hamiltonian and (6) Ham
ilton's equations in rectangular coordinates (x,y,z).
Ans. (a) H = (p + p\ + p)/2m + V(x, y, z)
(b) x = pjm, y = p y /m, z = p z /m, p x = dV/dx, p y = dV/dy, p z = dV/dz
12.28. Use Hamilton's equations to obtain the motion of a particle of mass m down a frictionless inclined
plane of angle a.
12.29. Work the problem of small oscillations of a simple pendulum by using Hamilton's equations.
12.30. Use Hamilton's equations to obtain the motion of a projectile launched with speed v at angle a
with the horizontal.
CHAP. 12]
HAMILTONIAN THEORY
333
12.31. Using Hamilton's equations, work the problem of the harmonic oscillator in (a) one dimension,
(b) two dimensions, (c) three dimensions.
12.32. Work Problem 3.27, page 78 by uging Hamilton's equation.
PHASE SPACE AND LIOUVILLE'S THEOREM
12.33. Explain why the path of a phase point in phase space which represents the motion of a system of
particles can never cross itself.
12.34. Carry out the details in the proof of Liouville's theorem for the case of two degrees of freedom.
CALCULUS OF VARIATIONS AND HAMILTON'S PRINCIPLE
12.35. Use the methods of the calculus of variations to find that curve connecting two fixed points in a
plane which has the shortest length.
12.36. Prove that if the function F in the integral J F(x, y, y') dx is independent of x, then the integral
a
is an extremum if F — y'F y > — c where c is a constant.
12.37. Use the result of Problem 12.36 to solve (a) Problem 12.9, page 322, (6) Problem 12.35.
12.38. It is desired to revolve the curve of Fig. 125 hav
ing endpoints fixed at P(x v y x ) and Q(x 2 , y 2 ) about
the x axis so that the area / of the surface of
revolution is a minimum.
/**
yy/l + y' % dx.
(b) Obtain the differential equation of the curve.
(c) Prove that the required curve is a catenary.
Ans. (6) yy" = 1 + (y')2
12.39. Two identical circular wires in contact are placed
in a soap solution and then separated so as to
form a soap film. Explain why the shape of the
soap film surface is related to the result of Prob
lem 12.38.
Q(x 2 , y 2 )
Fig. 125
12.40. Use Hamilton's principle to find the motion of a simple pendulum.
12.41. Work the problem of a projectile by using Hamilton's principle.
12.42. Use Hamilton's principle to find the motion of a solid cylinder rolling down an inclined plane of
angle a.
CANONICAL TRANSFORMATIONS AND GENERATING FUNCTIONS
12.43. Prove that the transformation Q = p, P = —q is canonical.
12.44. Prove that the transformation Q = q tan p, P = In sin p is canonical.
12.45. (a) Prove that the Hamiltonian for a harmonic oscillator can be written in the form H = 4p 2 /ra +
(b) Prove that the transformation q = \P/y/^sinQ, p  \j mPyf^ cos Q is canonical.
(c) Express the Hamiltonian of part (a) in terms of P and Q and show that Q is cyclic.
(d) Obtain the solution of the harmonic oscillator by using the above results.
334 HAMILTONIAN THEORY [CHAP. 12
12.46. Prove that the generating function giving rise to the canonical transformation in Problem 12.45(6)
is S — ^yficq 2 cotQ.
12.47. Prove that the result of two or more successive canonical transformations is also canonical.
12.48. Let V be a generating function dependent only on Q a ,p a ,t. Prove that
12.49. Let °L> be a generating function dependent only on the old and new momenta p a and P a respectively
and the time t. Prove that ^ w _ w
12.50. Prove that the generating function K of Problem 12.48 is related to the generating function T of
Problem 12.12 by V = T  2p«<7a
12.51. Prove that the generating function "V of Problem 12.49 is related to the generating function T
Problem 12.12 by "V = T + ^P a Q a  ^p a q a 
THE HAMILTONJACOBI EQUATION
12.52. Use the HamiltonJacobi method to determine the motion of a particle falling vertically in a uniform
gravitational field.
12.53. (a) Set up the HamiltonJacobi equation for the motion of a particle sliding down a frictionless
inclined plane of angle a. (b) Solve the HamiltonJacobi equation in (a) and thus determine the
motion of the particle.
12.54. Work the problem of a projectile launched with speed v Q at angle a with the horizontal by using
HamiltonJacobi methods.
12.55. Use HamiltonJacobi methods to describe the motion and find the frequencies of a harmonic oscil
lator in (a) 2 dimensions, (b) 3 dimensions.
12.56. Use HamiltonJacobi methods to arrive at the generating function of Problem 12.46.
PHASE INTEGRALS AND ANGLE VARIABLES
12.57. Use the method of phase integrals and angle variables to find the frequency of a simple pendulum
of length I, assuming that oscillations are small. An$. ^^j
12.58. Find the frequencies of (a) a 2 dimensional harmonic oscillator, (b) a 3 dimensional harmonic
oscillator.
12.59. Obtain the frequency of small oscillations of a compound pendulum by using phase integrals.
12.60. Two equal masses m connected by equal
springs to fixed walls at A and B are free
to slide in a line on a frictionless plane AB V A
to slide in a line on a irictioniess piane ad y, m m ft
[see Fig. 126]. Using phase integrals deter  —nS^^C\^W^C\ 'W L p
mine the freauencies of the normal modes. ^A ^ \^# vvv \l# £^
mine the frequencies of the normal modes. ^
1
12.61. Discuss Problem 12.57 if oscillations are not
assumed small. t<Ig
1
MISCELLANEOUS PROBLEMS
12 62 A particle of mass m moves in a force field having potential V( P ,*,z) where P ,*,z are cylindrical
coordinates. Give (a) the Hamiltonian and (b) Hamilton's equations for the particle.
Ans. (a) H = (p 2 p + p%h 2 + P 2 z )/2m + V( P , <p, z)
(b) P = Pp/m, I = Po/mr*. z = Vz /m, p p = p%lm P *  dV/d P , P<t> = dV/d<f>, p z = ~dV/dz
CHAP. 121
HAMILTONIAN THEORY
335
12.63.
12.64.
A particle of mass m which moves in a plane relative to a fixed set of axes has a Hamiltonian
given by the total energy. Find the Hamiltonian relative to a set of axes which rotates at constant
angular velocity «* relative to the fixed axes.
Set up the Hamiltonian for a double pendulum. Use HamiltonJacobi methods to determine the
normal frequencies for the case of small vibrations.
12.65.
X*2
F{t,x,x'x)dt to be an extremum is that
Bx dt \dx J + dV\dx) ~ °
12.66.
12.67.
BF d /BF
Can you generalize this result?
Work Problem 3.22, page 76, by Hamiltonian methods
A particle of mass m moves on the inside of a f rictionless vertical cone having equation x 2 + y^ =
* 2 tan 2 a. (a) Write the Hamiltonian and (b) Hamilton's equations using cylindrical coordinates.
Arts. (a) H = ^ — + ir ± s + mgp cot a
2m
(6) } = P " Sin2a
m
2m P 2
P<ft
rap 3
— mg cot a
12.68.
Use the results of Problem 12.67 to prove that there will be a stable orbit in any horizontal plane
z — h > 0, and find the frequency in this orbit.
12.69.
12.70.
Prove that the product of a position coordinate and its canonically conjugate momentum must
have the dimension of action or energy multiplied by time, i.e. AfL 2 Ti.
Perform the integration of equation (10) of Problem 12.16 and compare with the solution of the
Kepler problem in Chapter 5.
12.71.
12.72.
12.73.
Verify the integration result (3) of Problem 12.21.
Prove that Euler's equation (9), page 313, can be written as
.n^E., „, &F
+
dW
dF
12.74.
12.75.
V ' dy'2 + y dy'dy ' dy' dx by
A man can travel by boat with speed v 1 and can walk
with speed v 2 . Referring to Fig. 127, prove that in
order to travel from point A on one side of a river
bank to a point B on the other side in the least time
he must land his boat at point P where angles 9 t and
2 are such that
sm 0j _ Vj
sin e 2 v 2
Discuss the relationship of this result to the refrac
tion of light in the theory of optics.
Prove that if a particle moves under no external
forces, i.e. it is a free particle, then the principle of
least action becomes one of least time. Discuss the
relationship of this result to Problem 12.73. Fig. 127
Derive the condition for reflection of light in optical theory by using the principle of least time.
336
HAMILTONIAN THEORY
[CHAP. 12
12.76.
It is desired to find the shape of a curve lying in a plane and having fixed endpoints such that its
moment of inertia about an axis perpendicular to the plane and passing through a fixed origin is a
minimum.
(a) Using polar coordinates (r, e), show that the problem is equivalent to minimizing the integral
 J
r 2 V 1 + r 2 (de/dr) 2 dr
12.77.
12.78.
12.79.
where the fixed endpoints of the wire are (r v e^, (r 2 , 2 ).
(b) Write Euler's equation, thus obtaining the differential equation of the curve.
(c) Solve the differential equation obtained in (b) and thus find the equation of the curve.
Ans. (c) r 3 = c t sec (30 — c 2 ) where c t and c 2 are determined so that the curve passes through
the fixed points.
Use the HamiltonJacobi method to set up the equations of motion of a spherical pendulum.
Use HamiltonJacobi methods to solve Problems 11.20, page 293, and 11.21, page 294.
If [F,G] is the Poisson bracket [see Problems 12.25 and 12.26], prove that
(a) [F.F^G] = F t [F 2 ,G] + F 2 [F lt G\
<»> f t [F,G) =
(c)
A
dt
[F,G] =
dF
_dt
dF_
dt
,G
,G
+
+
w dG
F,
dG
dt
12.80.
12.81.
12.82.
12.83.
12.84.
12.85.
12.86.
12.87.
Prove that (a) [q a , q fi ] = 0, (6) [p a , p fi ] = 0, (c) [p a , q p ] = S a(3
fl if a = fi
where t afl = Q tf
is called the Kronecker delta.
Evaluate [H, t] where H is the Hamiltonian and t is the time. Are H and t canonically conjugate
variables? Explain.
Prove JacobVs identity for Poisson brackets
{F lt [F 2 ,F 8 ]] + \F 2 , [F a ,F^ + [F3,[F lf F 2 ]] =
Illustrate Liouville's theorem by using the one dimensional harmonic oscillator.
(a) Is the Lagrangian of a dynamical system unique? Explain.
(6) Discuss the uniqueness of the generalized momenta and Hamiltonian of a system.
(a) Set up the Hamiltonian for a string consisting of N particles [see Problem 8.29, page 215]
(6) Use HamiltonJacobi methods to find the normal modes and frequencies.
Prove that the Poisson bracket is invariant under a canonical transformation.
Prove that Liouville's theorem is equivalent to the result dp/dt = [p,H],
12.88.
12.89.
(a) Let Q a = 2 <*>anQ», p * = 2 *W?V where <W and 6 <*m are £ iven constants and
a = 1, 2, . . . , n. Prove that the transformation is canonical if and only if b ail — A a(l /A
where A is the determinant.
«11 a 12 • • • «*ln
a 21 a 22
a 2n
a n\ a n2
a„
and A ajLt is the cof actor of the element a aix in this determinant.
(6) Prove that the conditions in (a) are equivalent to the condition 2 P a Qa = 2 Pa 9a
Prove that the path taken by a fixed point on a circle as it rolls along a given line is a cycloid.
CHAP. 121
HAMILTONIAN THEORY 337
12.90. (a) Express as an integral the total potential energy of a uniform chain whose ends are suspended
from two fixed points. (6) Using the fact that for equilibrium the total potential energy is a mini
mum, use the calculus of variations to show that the equation of the curve in which the chain
hangs is a catenary as in Problem 7.32, page 186. [Hint. Find the minimum of the integral subject
to the constraint condition that the total length of the chain is a given constant.]
12.91. Use the methods of the calculus of variations to find the closed plane curve which encloses the
largest area.
12.92. Prove that the constants (a) /? in Problem 12.15 and (b) fo in Problem 12.16 can be identified with
the total energy.
12.93. If the theory of relativity is taken into account in the motion of a particle of mass m in a force
field of potential V, the Hamiltonian is given by
H = vVc 2 + m 2 c 4 + V
where c is the speed of light. Obtain the equations of motion for this particle.
12.94. Use Hamiltonian methods to solve the problem of a particle moving in an inverse cube force field.
12.95. Use spherical coordinates to solve Kepler's problem.
12.96. Suppose that m of the n coordinates q lt q 2 , . . . , q n are cyclic [say the first ra, i.e. q x , q 2 , . . . , q m ]' Let
TO
% — 2 c a h a ~ L where c a = dL/dq a
a=l
Prove that for a = m + 1, . . .,n 77 ( tj ) — t— 
dt \ dq a J dq a
The function % is called Routh's function or the Routhian. By using it a problem involving n
degrees of freedom is reduced to one involving n — m degrees of freedom.
rlT rlT
12.97. Using the properties SL = — 8y + ^7 8y', (8y)' = 8y'
of the variational symbol 8 [see Problem 12.6] and assuming that the operator 8 can be brought
under the integral sign, show how Lagrange's equations can be derived from Hamilton's principle.
12.98. Let P = Pip, q), Q = Q(p, q). Suppose that the Hamiltonian expressed in terms of p, q and P, Q
are given by H  H(p, q) and J[ = J((P, Q) respectively. Prove that if
q = 8H/dp, p = dH/dq
then Q = dJl/dP, P = dJ{/dQ
provided that the Jacobian determinant [or briefly Jacobian]
d(P, Q) _ dP/dp dP/dq
Hp,q) ~ dQ/dp dQ/dq
Discuss the connection of the results with Hamiltonian theory.
12.99. (a) Set up the Hamiltonian for a solid cylinder rolling down an inclined plane of angle a.
(b) Write Hamilton's equations and deduce the motion of the cylinder from them.
(c) Use HamiltonJacobi methods to obtain the motion of the cylinder and compare with part (6).
12.100. Work Problem 7.22, page 180, by HamiltonJacobi methods.
= 1
338 HAMILTONIAN THEORY [CHAP. 12
12.101. Write (a) the Hamiltonian and (6) Hamilton's equations for a particle of charge e and mass m
moving in an electromagnetic field [see Problem 11.90, page 309].
Ans. (a) H  ^(p eA) 2 + e<*>
(6) v = — (peA), p = eV* + eV(A«v)
12.102. (a) Obtain the HamiltonJacobi equation for the motion of the particle in Problem 12.101. (6) Use
the result to write equations for the motion of a charged particle in an electromagnetic field.
12.103. (a) Write the Hamiltonian for a symmetrical top and thus obtain the equations of motion. (6) Com
pare the results obtained in (a) with those of Chapter 10.
12.104. Prove Theorem 12.2, page 314.
12.105. An atom consists of an electron of charge — e moving in a central force field F about a nucleus of
F =
charge Ze such that _ 2
r3
where r is the position vector of the electron relative to the nucleus and Z is the atomic number.
In Bohr's quantum theory of the atom the phase integrals are integer multiples of Planck's con
stant h, i.e., ~ •»
<j) p r dr = n^h. d) p e de — n 2 h
Using these equations, prove that there will be only a discrete set of energies given by
„ 2?r 2 mZ 2 e 4
where n = n x + n 2 = 1, 2, 3, 4, . . . is called the orbital quantum number.
Appendix A
Units and Dimensions
UNITS
Standardized lengths, times and masses in terms of which other lengths, times and
masses are measured are called units. For example, a distance can be measured in terms of
a standard foot or meter. A time interval can be measured in terms of seconds; hours or
days. A mass can be measured in terms of pounds or grams. Many different types of units
are possible. However, there are four main types in use at the present time.
1. The CGS or centimetergramsecond system.
2. The MKS or meterkilogramsecond system.
3. The FPS or footpoundsecond system.
4. The FSS or footslugsecond system, also called the
English gravitational or engineering system.
The first two are sometimes called metric systems, while the last two are sometimes called
English systems. There is an increasing tendency to use metric systems.
The following indicates four consistent sets of units in these systems which can be used
with the equation F = ma:
CGS System: F (dynes) = m (grams) x a (cm/sec 2 )
MKS System: F (newtons) = m (kilograms) x a (m/sec 2 )
FPS System: F (poundals) = m (pounds) x a (ft/sec 2 )
FSS System: F (pounds weight) = m (slugs) x a (ft/sec 2 )
In the third through sixth columns of the table on page 340, units of various quantities in
these systems are given.
In the table on page 341, conversion factors among units of the various systems are
given.
DIMENSIONS
The dimensions of all mechanical quantities may be expressed in terms of the funda
mental dimensions of length L, mass M, and time T. In the second column of the table on
page 340, the dimensions of various physical quantities are listed.
339
340
UNITS AND DIMENSIONS
[APPENDIX A
UNITS AND DIMENSIONS
Physical Quantity
Dimension
CGS System
MKS System
FPS System
FSS System
Length
L
cm
m
ft
ft
Mass
M
gm
kg
lb
slug
Time
T
sec
sec
sec
sec
Velocity
LTi
cm/sec
m/sec
ft/sec
ft/sec
Acceleration
LT*
cm/ sec 2
m/sec 2
ft/sec 2
ft/ sec 2
Force
MLT*
gm cm/sec 2
= dyne
kg m/sec 2
= newton
lb ft/sec 2
= poundal
slug ft/sec 2
= lbwt
Momentum, Impulse
MLTi
gm cm/sec
= dyne sec
kg m/sec
= nt sec
lb ft/sec
= pdl sec
slug ft/ sec
= lbwt sec
Energy, Work
ml* r2
gm cm 2 / sec 2
= dyne cm
= erg
kg m 2 /sec 2
= nt m
= joule
lb f t 2 /sec 2
= ft pdl
slug ft 2 /sec 2
= ft lbwt
Power
ML 2 ra
gm cm 2 /sec 3
= dyne cm/sec
= erg/sec
kg m 2 /sec 3
= joule/sec
= watt
lb f t 2 /sec 3
= ft pdl/sec
slug f t 2 /sec 3
= ft lbwt/sec
Volume
L 3
cm 3
m 3
ft 3
ft 3
Density
ML3
gm/cm 3
kg/m 3
lb/ft 3
slug/ft 3
Angle
—
radian (rad)
rad
rad
rad
Angular velocity
T i
rad/sec
rad/sec
rad/sec
rad/sec
Angular acceleration
ji2
rad/sec 2
rad/sec 2
rad/sec 2
rad/sec 2
Torque
ML 2 T2
gm cm 2 /sec 2
= dyne cm
kg m 2 /sec 2
= nt m
lb ft 2 /sec 2
= ft pdl
slug ft 2 /sec 2
= ft lbwt
Angular momentum
ML 2 T~i
gm cm 2 /sec
kg m 2 /sec
lb f Wsec
slug ft 2 /sec
Moment of inertia
Ml?
gm cm 2
kg m 2
lb ft 2
slug ft 2
Pressure
MLi T2
gm/(cm sec 2 )
= dyne/cm 2
kg/(m sec 2 )
= nt/m 2
pdl/ft 2
lbwt/ft 2
J .
APPENDIX A]
UNITS AND DIMENSIONS
341
CONVERSION FACTORS
Length
Area
Volume
Mass
Speed
Density
Force
Energy
Power
Pressure
1 kilometer (km)
1 meter (m)
1 centimeter (cm)
1 millimeter (mm)
1 micron (/t)
1 millimicron (m/t)
1 angstrom (A)
1000 meters
100 centimeters
102 m
10" 3 m
10" 6 m
109 m
10 10 m
1 inch (in.)
1 foot (ft)
1 mile (mi)
lmil
1 centimeter
1 meter
1 kilometer
2.540 cm
30.48 cm
1.609 km
10 3 in.
0.3937 in.
39.37 in.
0.6214 mile
1 square meter (m 2 ) = 10.76 ft 2
1 square foot (ft 2 ) = 929 cm 2
1 square mile (mi 2 ) = 640 acres
1 acre = 43,560 ft 2
1 liter (I) = 1000 cm 3 = 1.057 quart (qt) = 61.02 in 3 = 0.03532 ft 3
1 cubic meter (m 3 ) = 1000 I = 35.32 ft 3
1 cubic foot (ft 3 ) = 7.481 U.S. gal = 0.02832 m 3 = 28.32 I
1 U.S. gallon (gal) = 231 in 3 = 3.785 I; 1 British gallon = 1.201 U.S. gallon = 277.4 in 3
1 kilogram (kg) = 2.2046 lb = 0.06852 slug; 1 lb = 453.6 gm = 0.03108 slug
1 slug = 32.174 lb = 14.59 kg
1 km/hr = 0.2778 m/sec = 0.6214 mi/hr = 0.9113 ft/sec
1 mi/hr = 1.467 ft/sec = 1.609 km/hr = 0.4470 m/sec
1 gm/cm 3 = 10 3 kg/m 3 = 62.43 lb/ft 3 = 1.940 slug/ft 3
1 lb/ft 3 = 0.01602 gm/cm 3 ; 1 slug/ft 3 = 0.5154 gm/cm 3
1 newton (nt) = 10 5 dynes = 0.1020 kgwt = 0.2248 lbwt
1 pound weight (lbwt) = 4.448 nt = 0.4536 kgwt = 32.17 poundals
1 kilogram weight (kgwt) = 2.205 lbwt = 9.807 nt
1 U.S. short ton = 2000 lbwt; 1 long ton = 2240 lbwt; 1 metric ton = 2205 lbwt
1 joule = 1 nt m = 10 7 ergs = 0.7376 ft lbwt = 0.2389 cal = 9.481 X 10 " 4 Btu
1 ft lbwt = 1.356 joules = 0.3239 cal = 1.285 X 10" 3 Btu
1 calorie (cal) = 4.186 joules = 3.087 ft lbwt = 3.968 X 10~ 3 Btu
1 Btu = 778 ft lbwt = 1055 joules = 0.293 watt hr
1 kilowatt hour (kw hr) = 3.60 X 10 6 joules = 860.0 kcal = 3413 Btu
1 electron volt (ev) = 1.602 X 10~ 19 joule
1 watt = 1 joule/sec = 10 7 ergs/sec = 0.2389 cal/sec
1 horsepower (hp) = 550 ft lbwt/sec = 33,000 ft lbwt/min = 745.7 watts
1 kilowatt (kw) = 1.341 hp = 737.6 ft lbwt/sec = 0.9483 Btu/sec
1 nt/m 2 = 10 dynes/cm 2 = 9.869 X lO" 6 atmosphere = 2.089 X 10~ 2 lbwt/ft 2
1 lbwt/in 2 = 6895 nt/m 2 = 5.171 cm mercury = 27.68 in. water
1 atmosphere (atm) = 1.013 X 10 5 nt/m 2 = 1.013 X 10 6 dynes/cm 2 = 14.70 lbwt/in 2
= 76 cm mercury = 406.8 in. water
Angle
1 radian (rad) = 57.296°; 1° = 0.017453 rad
Appendix B
Astronomical Data
Mass
Radius
Mean density
Mean surface gravitational
acceleration
Escape velocity
at surface
Period of rotation
about axis
Universal gravitational
constant G
THE SUN
4.4 X 10 30 lb or 2.0 X 10 30 kg
4.32 X 10 5 mi or 6.96 X 10 5 km
89.2 lb/ft 3 or 1.42 gm/cm 3
896 ft/sec 2 or 273 m/sec 2
385 mi/sec or 620 km/sec
25.38 days or 2.187 X 10 6 sec
1.068 X 10~ 9
ft 3 /lbsec 2
or 6.673 X 10 ~ 8 cm 3 /gmsec 2
Mean distance from earth
Period of rotation
about earth
Equatorial radius
Mass
Mean density
Mean surface gravitational
acceleration
Escape velocity
Period of rotation
about axis
Orbital speed
Orbital eccentricity
THE MOON
239 X 10 3 mi or 3.84 X 10 5 km
27.3 days or 2.36 X 10 6 sec
1080 mi or 1738 km
1.63 X 10 23 lb or 7.38 X 10 22 kg
27.3 days
.055
208 lb/ft 3 or 3.34 gm/cm 3
5.30 ft/sec 2 or 1.62 m/sec 2
1.48 mi/sec or 2.38 km/sec
or 2.36 X 10 6 sec
.64 mi/sec or 1.02 km/sec
342
APPENDIX Bl
ASTRONOMICAL DATA
343
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Appendix C
Solutions of Special
Differential Equations
DIFFERENTIAL EQUATIONS
An equation which has derivatives or differentials of an unknown function is called a
differential equation. The order of the differential equation is the order of the highest
derivative or differential which is present. A solution of a differential equation is any
relationship between the variables which reduces the differential equation to an identity.
Example 1.
The equation j*=2y is a differential equation of first order, or order 1. A solution of this equation
is y = ce 2x where c is any constant, since on substituting this into the given differential equation we have
the identity
2ce 2x = 2ce 2x
Example 2.
The equation x 2 dx + y 3 dy = is a differential equation of first order. A solution is x*/3 + y*l± = c
where c is any constant, since taking the differential of the solution we have
d(x s /S + y*/4) = or x 2 dx + y z dy 
Example 3.
The equation ^  3^ + 2y = 4x is a differential equation of second order. A solution is
CLOG 0/QCr
y = de x + c 2 e 2x + 2x + 3 since
&£ S *!L + 2 y = ( Cl e x + 4c 2 e 2x )  3( Cl e x + 2c 2 e 2x + 2) + 2(c t e x + c 2 e 2x + 2x + 3)  4*
dx 2 dx
In the above examples we have used x as independent variable and y as dependent
variable. However, it is clear that any other symbols could just as well have been used.
Thus, for instance, the differential equation of Example 3 could be
H  3^ + 2x = U
dt 2 dt
with independent variable t and dependent variable * and solution x = cie* + c 2 e 2t + 2t + S.
The above equations are often called ordinary differential equations to distinguish them
a2V r) 2 V
from partial differential equations such as ^ = c 2 ^" involving two or more independent
variables.
ARBITRARY CONSTANTS. GENERAL AND PARTICULAR SOLUTIONS
In the above examples the constants c, d, c 2 can take on any values and are called
arbitrary constants. In practice an nth order differential equation will have a solution
involving exactly n independent arbitrary constants. Such a solution is called the general
solution. All special cases of the general solution obtained by giving the constants special
values are then called particular solutions. For instance in Example 3 above if we let
ci = 6, c 2 = 3 in the general solution y = w* + c 2 e 2 * + 2x + 3, we obtain the particular
solution y = 5e x  3e 2x + 2x + 3.
344
APPENDIX C] SOLUTIONS OF SPECIAL DIFFERENTIAL EQUATIONS 345
Particular solutions are often found from certain conditions imposed on the problem
and sometimes called boundary or initial conditions. In Example 3 for instance, if we wish
to satisfy the conditions y = 5 when x = and if = dy/dx = 1 when x = 0, we obtain
ex = 5, c 2 = —3.
A problem in which we are required to solve a differential equation subject to given
conditions is often called a boundaryvalue problem.
SOLUTIONS TO SOME SPECIAL FIRST ORDER EQUATIONS
The following list shows some important methods for finding general solutions of first
order differential equations.
1. Separation of Variables
If a first order equation can be written as
F(x)dx + G(y)dy = (1)
then the variables are said to be separable and the general solution obtained by direct
integration is
J F(x) dx + C G(y) dy =■ c (2)
2. Linear Equations
A first order equation is called linear if it has the form
g + P(x)y = Q(x) (S)
Multiplying both sides by e J , this can be written
^{yJ Pdx } = Qe Spdx
Then integrating, the general solution is
y e ) pdx — 1 Qe^ Pdx dx + c
or y = e~f Pdx C Qef Pdx dx + ceS Pdx (4)
C p dx
The factor e J is often called an integrating factor.
3. Exact Equation
The equation
Mdx + Ndy = (5)
where M and N are functions of x and y is called an exact differential equation if
Mdx + Ndy can be expressed as an exact differential dU of a function U(x,y). In such
case the solution is given by U(x, y) = c.
A necessary and sufficient condition that (5) be exact is
§M = dN
By dx K '
In some cases an equation is not exact but can be made exact by first multiplying
through by a suitably chosen function called an integrating factor as in the case of the
linear equation.
346 SOLUTIONS OF SPECIAL DIFFERENTIAL EQUATIONS [APPENDIX C
4. Homogeneous Equation
If an equation has the form
P = f(V) (7)
dx \x/
it is called a homogeneous equation and can be solved by the transformation y = vx.
Using this, (7) becomes
dv r,/ v dx dv , s ,
v + x Tx = F{v) or ~x~ = fW 1 ^ {)
in which the variables have been separated. Then the general solution is
fdx = C dv c where v = /x {9)
J x J F(v)  v
Occasionally other transformations, which may or may not be evident from the form
of a given differential equation, serve to obtain the general solution.
SOLUTIONS OF HIGHER ORDER EQUATIONS
The following list shows certain equations of order higher than one which can often
be solved.
1. ^ = F(x). In this case the equation can be integrated n times to obtain
dx n v '
y = f • • • f F(x) dx n + ci + c 2 x + czx 2 + ••• + e n x n ~ l
2 SlL = f ( x !^L) . in this case y is missing and if we make the substitution dy/dx = v
dx 2 \ dx J
we find that the equation becomes
dv „. x
Tx = F(x ' v)
a first order equation. If this can be solved we replace v by dy/dx, obtaining another
first order equation which then needs to be solved.
q ^y = p( v ^}L\ . Here x is missing and if we make the substitution dy/dx = v,
' dx 2 \ a 'dxJ
noting also that TO , , ,
d 2 y _ cLv_ _ dv&n _ dv
dx 2 ~ dx ~ dydx dy'
the given equation can be written as a first order equation
which then needs to be solved.
LINEAR DIFFERENTIAL EQUATIONS OF ORDER HIGHER THAN ONE
We shall consider solutions of linear second order differential equations. The results
can easily be extended to linear higher order equations.
A linear second order equation has the form
^ + P (x) f x + Q( X )y = IHx) (10)
APPENDIX C] SOLUTIONS OF SPECIAL DIFFERENTIAL EQUATIONS 347
If y c is the general solution of the equation
[obtained by replacing the right hand side of (10) by zero] and if y p is any particular solution
of (10), then the general solution of (10) is
y = y c + y p (12)
The equation (11) is often called the complementary equation and its general solution is
called the complementary solution.
LINEAR EQUATIONS WITH CONSTANT COEFFICIENTS
The complementary solution is easily obtained when P(x) and Q(x) are constants A and
B respectively. In such case equation (11) can be written
+ 4f + *" = » W
If we assume as solution y = e ax where a is constant in (11), we find that a must satisfy
the equation
a 2 + Aa + B =
This equation has two roots, and the following cases arise.
1. Roots are real and distinct, say a x ¥ a 2 .
In this case solutions are e aiX and e 012 *. It also follows that c x e a & and c 2 e a * x are solu
tions and that the sum c x e a & + c 2 e a * x is the general solution.
2. Roots are real and equal, say « 1 = « 2 .
In this case we find that solutions are e aiX and xe aiX and the general solution is
c x e a ^ x + c 2 xe aiX .
3. Roots are complex.
If A and B are real, these complex roots are conjugate, i.e. a + bi and a — bi. In such
case solutions are e ia+hi)x = e ax e bix = e ax (cos bx + i sin bx) and e (a ~ bi)x (cos bx — i sin bx).
The general solution can be written e ax (c x cos bx + c 2 sin bx).
PARTICULAR SOLUTIONS
To find the general solution of
we must find a particular solution of this equation and add it to the general solution of (13)
already obtained above. Two important methods serve to accomplish this.
1. Method of Undetermined Coefficients.
This method can only be used for special functions R(x) such as polynomials and the
exponential or trigonometric functions having the form e px , cos px, sin px where p
is a constant, together with sums and products of such functions. See Problems C.17
and C.18.
2. Method of Variation of Parameters.
In this method we first write the complementary solution in terms of the constants
c x and c 2 . We then replace c x and c 2 by functions f t (x) and f 2 (x) so chosen as to satisfy
the given equation. The method is illustrated in Problem C.19.
348 SOLUTIONS OF SPECIAL DIFFERENTIAL EQUATIONS [APPENDIX C
Solved Problems
DIFFERENTIAL EQUATIONS. ARBITRARY CONSTANTS.
GENERAL AND PARTICULAR SOLUTIONS
Cl. (a) Prove that y = ce~ x + x1 is the general solution of the differential equation
% ~ x+v = °
(b) Find the particular solution such that y = 3 when x  0.
(a) If y  ce~ x + x  1, then dy/dx = —ee~ x + 1 and so
dy/dx  x + y  (ce' x + 1)  x + (ce~ x + x  1) =
Thus 2/ = ce x + a; — 1 is a solution; and since it has a number of arbitrary constants
(namely one) equal to the order of the differential equation, it is the general solution.
(6) Since y = 3 when x = 0, we have from y = ce~ x + x1, 3 = c  1 or c = 4. Thus
y = 4e~ x + x — 1 is the required particular solution.
C.2. (a) Prove that x = c^ + c 2 e~ st + sin * is the general solution of
^ + 2^  3* = 2cos*  4 sin*
dt 2 at
(b) Find the particular solution such that x1, dx/dt = 3 at t = 0.
(a) From a = Cje* + c 2 e~ 3t + sin t we have
^ = Cl e*  3e 2 e 8t + cost, "^Jf = c i et + 9c 2 e 3t  sin t
Then ^ + 2 —  3a; = (c ie t + 9c 2 e3t _ sin t ) + 2( Cl e*  3c 2 e"« + cos t)
dt2 dt  3(cje* + c 2 e3t + sin t)
= 2 cos t — 4 sin t
Thus a; = c^* + c 2 e~ 3t + sin t is a solution; and since it has two arbitrary constants while the
differential equation is of order two, it is the general solution.
(6) From part (a), letting t = in the expressions for x and dx/dt, we have
2 = c t + c 2 fci + c 2 = 2
or i .
3 = Cl  3c 2 + 1 L c i _ 3c 2 = _4
Solving, we find c x — 1/2, c 2 = 3/2. Then the required particular solution is
x = ^e* + f e~ 3t + sin t
SEPARATION OF VARIABLES
C.3. (a) Find the general solution of (x + xy 2 ) dx + (y + £ 2 2/) dy = 0.
(b) Find the particular solution such that y = 2 when a; = 1.
(a) Write the equation as x(l + y 2 ) dx + y(l + x 2 ) dy = 0. Dividing by (1 + x 2 )(l + y 2 ) ¥> to
separate the variables, we find
xdx ydy _ Q ^)
1 + a;2 ^ 1 + y2
Then we have on integrating,
J xdx , f ydy
1 + x 2 J 1 + y
1 In (1 + a; 2 ) + £ In (1 + ?/ 2 ) = *i
^y* = Cl
APPENDIX C] SOLUTIONS OF SPECIAL DIFFERENTIAL EQUATIONS 349
This can be written ^ In {(1 + x 2 )(l + y 2 )}  c x or
(1 + X 2 ){1 + yl) = C2 (j g)
which is the required general solution.
(6) Since y = 2 when x = 1, we have on substitution in (2), c 2 = 10; thus the required particular
solution is
(1 + x 2 )(l + y 2 ) = 10 or x 2 + y 2 + x 2 y 2 = 9
C.4. Solve ^ = RH 2 if R = 1 when t = 1.
at
dR
Separating the variables, we have ^  t 2 dt. Integrating both sides,
1 t 3
~R = 3 +C
Substituting t = 1, R = 1 we find c = 4/3. Thus
1 t 3 4 „ 3
— or it
R 3 3 4  t3
LINEAR EQUATION
dv
C.5. Solve 5— + 2xy = x 3 + x if 2/ = 2 when x = 0.
This is a linear equation of the form (S), page 345, with P = 2x, Q = X s + x. An integrating
I 2.x dix 2
factor is e J — e x . Multiplying the given equation by this factor, we find
e x2 ^ + 2xye x2 = (x 3 + x)e x2
ax
j
which can be written t~{v e x2 ) — (# 3 + x)e x2
ax
Integrating, y e x = I (x 3 + x)e x2 dx + c
or, making the substitution v = x 2 in the integral,
y e* 2 = 1x20*2 + c
Thus y = ±x 2 + ce~ x2
Since y = 2 when x = 0, we find c = 2. Thus
y = \x 2 + 2e~ x2
Check: If y = \x 2 + 2e~ x2 , then dy/dx = x  4xe~ x2 . Thus
^+ 2xy = x  4xe~ x2 + 2x(\x 2 + 2e~* 2 ) = x 3 + x
C.6. Solve ~=SU + lifU = when t = 0.
dt
Writing the equation in the form
dU
dt~* U = 1 W
350 SOLUTIONS OF SPECIAL DIFFERENTIAL EQUATIONS [APPENDIX C
we see that it is linear with integrating factor e* ~ 3dt = e~ st . Multiplying (1) by e" 8 ', it can be
written as
4i(UeM) = e *t
at
Integrating, we have Ue~ 3t = — ^e~ 3t + c
Since U = when t = 0, we obtain c = £. Thus
Uezt = _£ e 3t + . or u = £(e 3t l) (2)
Another method. The equation can also be solved by the method of separation of variables. Thus
we have , r .
dU = dt
3U + 1
Integrating, £ In (3t7 + 1) = t + c
Since U = when t = 0, we find c = so that £ In (BU + 1) = t. Thus *7 = £(e 3t l).
EXACT EQUATIONS
C.7. Solve (Sx 2 + y cos x) dx + (sin x  4y 3 ) dy = 0.
Comparing with M dx + N dy = 0, we have M = Sx 2 + y cos x, N = sin g  4s/ 3 . Then
3Af/3i/ = cos x = dN/dx and so the equation is exact. Two methods of solution are available.
Method 1.
Since the equation is exact, the left side must be an exact differential of a function U. By
grouping terms, we find that the equation can be written
3a 2 dx + (y cos x dx + sin x dy) — Ay 3 dy =
i.e., d(x s ) + d(y sin *) + d(y±) = or d(» 3 + y sin x  y 4 ) =
Integration then gives the required solution, X s + y sin x — y 4 = c.
Method 2.
The given equation can be written as
(3a; 2 + y cos x) dx + (sin x  4j/ 3 ) dy = dU = ^dx + —dy
Then we must have
ATT STJ
(1) ^  3a;2 + 2/ cos a; (2) — = sin x  4j/ 3
v ' dx oy
Integrating (1) with respect to x, keeping y constant, we have
U = X s + y sin x + F(y)
Then substituting this into (2), we find
sin a; + F'(y) = sin x  4?/ 3 or F'(y) = 4^
where F'(y) = dF/dy. Integrating, omitting the constant of integration, we have F(y) = y* so
that
U = X s + y sin x — y*
Then the given differential equation can be written
dU = d(x* + y sin x — y 4 ) =
and so the solution is x z + y sin x — y 4 = c.
APPENDIX C] SOLUTIONS OF SPECIAL DIFFERENTIAL EQUATIONS 351
HOMOGENEOUS EQUATIONS
C.8. Solve p = e» /x + %.
dx x
Let y = vx. Then the equation can be written
dv , dv
v + %7 = e v + v or %~j~ — e
dx Ax
Separating the variables, — = e~ v dv. Integrating, Ins = — e~ v + c. Thus the general solution is
lnsc + e~ y,x — c.
SOLUTIONS OF HIGHER ORDER EQUATIONS
C.9. Solve ^ = 1 + cos t where U = 2, dU/dt = 3 at t = 0.
Integrating once,
dU/dt = t + sin* + c x
Then since dU/dt = 3 at * = 0, we find e t  3. Thus
dU/dt = t + sin* + 3
Integrating again, U = ^t 2 — cos t + St + c 2
Now since U = 2 at t  0, we find c 2 = 3. The required solution is
U = \&  cos t + 3t + 3
CIO. Solve a#" + 2y' = x 2 where y' = dy/dx, y" = d?y/dx 2 .
Since y is missi'ig, let y' = dy/da; = v. Then the equation can be written
(i) *g+2* = * 2 or (f) £+§* = x
in i> with integrating factor J (2/x) dx = e 2 ln * = c ln * 2 = * 2 . Multiplying
A
dx
Then by integration, a5 2 v = x*/4 + c x or
v = dy/dx = x*/4 + cjx*
Integrating again, y = a^/12 — c x lx + c 2 .
Cll. Solve yy" + {y'f = where y' = dy/dx, y"  d 2 y/dx 2 .
Since x is missing, let y' ~ dy/dx = v. Then
„ _ dv __ dv^djt _ v dv^
V ~ dx dy dx dy
and the given equation can be written
yv ^ + v2 = ° or v ( v % + v ) = °
/ x dv , _
so that (1) v = or (2) y^ + v =
From (1), y'=0 or 2/ = ^. From (2), — + ^ = 0, i.e. lnv + lny = c 2 or ln (vy) = c 2 so
that vy = c 3 and
This is a linear equation
(2) by 05 2 , it can be written as
(x 2 v) = x
352 SOLUTIONS OF SPECIAL DIFFERENTIAL EQUATIONS [APPENDIX C
v = dy/dx — c 3 /y or y dy = c 3 dx
Integrating, y 2 /2 = c 3 x + c 4 or y 2 = Ax + B
Thus solutions are y = c t and y 2 — Ax + B. Since the first is a special case of the second, the
required general solution can be written y 2 = Ax + B.
LINEAR EQUATIONS WITH CONSTANT COEFFICIENTS
C.12. Solve 4^4^52/ = 0.
dx 2 dx
Letting y = e ax in the equation, we obtain
(a 2  4a — h)e ax = or a 2 — 4a  5 =
Thus (a — 5)(a + 1) = and a = 5,1. Then solutions are e 5x and e~ x and the general solution is
y = c x e* x + c 2 e~ x .
C.13. Solve 0+10^ + 252/ = 0.
Letting y = e ax , we find a 2 + 10a + 25 = 0, i.e. (a + 5)(a + 5) = 0, or a = 5, 5. Since
the root is repeated, solutions are e~ 5x and xe~ 5x . Then the general solution is y = c x e~^ x + c 2 xe~ 5x .
C.14. Solve ^§ + 4^ + 4x = 0.
at 2 dt
Letting x = e at , we find a 2 + 4a + 4 = or a = 2, 2. Then the general solution is
x = c^" 2 * + c 2 te~ 2t = «~ 2t («i + c 2 t).
C.15. Solve + 2 g + 5v = 0.
Letting 2/ = e ax , we find a 2 + 2a + 5 = or a = 1 ± 2t. Then solutions are e (_1 + 2i)x 
e x e 2ix = e x( cos 2a; + t sin2a;) and e (1  2i)a; = e a: e 2te = ez(cos2ar — i sin2a;). The general
solution is y = e~ x (c x cos 2x + c 2 sin2x).
C.16. Solve d 2 y/dx 2 + <* 2 y  0.
Letting y = e ax , we find a 2 + w 2 = or a = ± ia. Then solutions are e to * = cos wo; + i sin w«
and e i6U! = cos <*x — i sin ax. The general solution is thus y = c t cos ax + c 2 sin v>x.
METHOD OF UNDETERMINED COEFFICIENTS
C.17. Solve 045i/ = x 2 + 2e*
By Problem C.12 the complementary solution, i.e. the general solution, of
&«£* = °
is y c = c t e& + c 2 e~ x (1)
Since the right side of the given equation contains a polynomial of the second degree (i.e. x 2 )
and an exponential (2e 3x ), we are led to the trial particular solution
y p = Ax 2 + Bx + C + De* x (*)
where A, B, C, D are constants to be determined.
Substituting (2) for y in the given equation and simplifying, we find
(2A  4B  BO + (8A  5B)x  SAs 2  8De 3 * = a 2 + 2e^ x
Since this must be an identity, we must have
APPENDIX C] SOLUTIONS OF SPECIAL DIFFERENTIAL EQUATIONS 353
2A4B5C = 0, 8A  55 = 0, 5A = 1, 8D = 2
Solving, we find A = £, B=£,C = {%, Z> = j. Then from (2),
2/p — 5 X T 25 X 125 4 e
Thus the required general solution of the given equation is
which can be checked by direct substitution.
C.18. Solve 0+lO^ + 252/ = 20 cos 2x.
The complementary solution [by Problem C.13] is
y c  Cl e 5x + c 2 xe 5x CO
Since the right side has the term cos 2x, we are led to the trial solution
y p = A cos 2x + B sin 2x (2)
Substitution into the given equation yields, after simplifying,
(21A + 205) cos 2x + (215  20A) sin 2x = 20 cos 2x
Equating coefficients of like terms, we have 21A + 205 = 20, 215  20/1 = 0. Solving, we find
A = 84i> B = 84i so that the P articular solution is
420 n i 400 . o
y v = I_t cos 2* + 57T sin 2a
'P 841
and the general solution of the given equation is
V = Vc + V P = c i e ~ Sx + We~s* + fff cos 2x +
420 ___ o„ , 400
841
sin 2x
METHOD OF VARIATION OF PARAMETERS
C.19. Solve d 2 y/dx 2 + y = tana:.
The complementary solution is as in Problem C.16 with u = 1:
y c = c_ cos x + c 2 sin a CO
We now assume that the solution to the given equation has the form
y = fi cos x + / 2 sin x (2)
where /_ and / 2 are suitable functions of a;. From (*) we have, using primes to denote differentiation
with respect to x,
dy/dx = fi sin * + f 2 cos x + /_ cos * + / 2 sin (5)
Before finding d 2 y/dx 2 let us observe that since there are two functions / t and / 2 to be determined
and only one relation to be satisfied [namely that the given differential equation must be satisfied]
we are free to impose one relation between f x and / 2 . We choose the relation
/i cos x + / 2 sin x = (4)
so as to simplify (8) which then becomes
dy/dx = — /1 sin x + f 2 cos a (5)
Another differentiation then leads to
d 2 y/dx 2 = fi cos *  / 2 sin x  f[ sin x + / 2 cos as (0)
From (2) and (0) we see that the given differential equation can be written
dty/dx* + y  fi sin x + f 2 cos x = tan x (7)
Thus — /{ sin x + / 2 cos x = tan x (5)
From (.4) and (5) we find fi =  sin 2 x/cos x, / 2 = sin x. Thus
354 SOLUTIONS OF SPECIAL DIFFERENTIAL EQUATIONS [APPENDIX C
C sin 2 x , f 1  cos 2 x , C . , ,
f. = — I ax = — I ax = — I (sec * — cos x) ax
J1 J cos* J cos x J
= — In (sec x + tan a;) + sin a? + c x
f 2 = \ sin x dx = — cos * + c 2
Substituting in (2) we find the required general solution,
y = d cos x + c 2 sin a; — cos a; In (sec x + tan x)
Supplementary Problems
DIFFERENTIAL EQUATIONS. ARBITRARY CONSTANTS.
GENERAL AND PARTICULAR SOLUTIONS
C.20. Check whether each differential equation has the indicated general solution.
(a) ^f  2^ + x = t ; x = (c t + c 2 t)e* + t + 2
at' at
(*>) *4r + V = &; t?StU = c
at
C.21. (a) Show that z = e _t (Ci sin t + c 2 cost) is a general solution of
(6) Determine the particular solution such that z = — 2 and dz/dt = 5 at t = 0.
Ans. (6) « = e t (3 sin * — 2 cos t)
SOLUTIONS TO SPECIAL FIRST ORDER EQUATIONS
C.22. Solve dy/dx = 2xy if y = 4 when x = 0. An*. # = 4e a:2
C.23. Solve
dz = ty/T^z* A ^^—^_^ r —
dt
Vi  * 2
C.24. Solve x^2y = x if y(l) = 5. Arts, y = 6a; 2  x
C.25. Solve (a; + 2y) dx + (2a;  hy) dy = if 1/ = 1 when a; = 2. Ans. a; 2 + 4»y  5y 2  7
C.26. Solve ^ =  + ^a Ans. \nx + (x/y) = c
dx x x z
C.27. Solve (ye x — e~v) dx + (xe~y + e x ) dy = 0. Ans. ye x — xe~ y  c
C.28. Solve (x + xy) dx + (xy + y)dy = 0. Ans. (x + l)(y + 1) = ce* + y
C.29. Show that the differential equation {Ay  a; 2 ) dx + xdy = has an integrating factor which de
pends on only one variable and thus solve the equation. Ans. xty — ^x 6 = c
SOLUTIONS OF HIGHER ORDER EQUATIONS
C.30. Solve d?U/dt z = t + e~* if U = 3, dU/dt = 2 at t = 0. Ans. U = ±t? + e~* + 3t + 2
C.31. Solve x J^  3 ^ = a; 2 . Ans. y = ^x» + c x x* + c 2
APPENDIX C] SOLUTIONS OF SPECIAL DIFFERENTIAL EQUATIONS 355
C.32. Solve ^ff+ 2 (^) 2 = °' Ans. US = Cl t + c 2
C.33. Solve
H*)"]' (£)'• ^<*^ +<**>* = !
LINEAR HIGHER ORDER DIFFERENTIAL EQUATIONS
C.34. Solve ^2^82/ = 0. Arts, y = c x e* x + c^ 2 *
C.35. Solve ^? + 4^+ W = if U = l,dU/dt = at t = 0. Ans. U = (l + 2t)e 2 *
dt £ at
C.36. Solve ^ + 4^J+ 5z = 0. Ans. z = e~ 2t (ci cos t + c 2 sin*)
ac z at
C.37. Solve 4^ + 25y = if 2/(0) = 10, y'(0) = 25. Ans. y = 10(cos§05 + sulfa)
C.38. Solve042/ = 0. Ans. y = e 2x ( Cl e^ x + c 2 e~^ x )
C.39. Solve 4y"  20y' + 25y = 0. Ans. (c t + c 2 x)e 5x/2
C.40. Solve + 3^ + 2y = ««*. Ans. y = Cl e~* + c 2 e~^ + \e~* x
C.41. Solve ^ + ^r~ 2U = 6t  10 cos 2* + 5.
at 2 at
Ans. U = c x e 2t + c 2 e*  St  4 +  sin 2t  § cos 2t
C.42. Solve y" + y — sec « by use of the method of variation of parameters.
Ans. y = c x sin x + c 2 cos x + x sin a; — cos a; In sec x
C.43. Solve (a) Problem C.40 and (6) Problem C.41 by variation of parameters.
C.44. Solve each of the following equations by any method.
(a) y"  5y' + 6y = 50 sin 4a; (c) y" + 4y = esc 2*
(6) y" + 2y'  3y = xe~ x (d) 1/" + 82/' + 25y = 25x + 33 + 18e"*
A«s. (a) y = c x e 2x + c 2 e Sx + 2 cos 4* — sin 4a?
(6) ^ = Cl e x + c 2 e~ Sx — \xe~ x
(c) y = c t cos 2x + c 2 sin 2a;  %x cos 2x — \ sin 2a; In (esc 2a:)
(^ y = e 4x( Cl cos 3* + c 2 sin 3a;) + x + 1 + e _x
C.45. Solve y" + 4j/' + 4y = e" 2 *. Aw*, y = (e t + c 2 a;)e~ 2a; + \xH~^
C.46. Solve simultaneously: dx/dt + y = e*, x — dy/dt = t.
Ans. x = c t cos t — c 2 sin t + ^e* + t,
y = d sin t + c 2 cos £ + £e* — 1
C.47. Solve 2/" + 2/ = 4 cos t if 2/(0) = 2, y'(0) ~ 1. Is the method of undetermined coefficients ap
plicable? Explain. Ans. y = 2 cos t  sin t + 2t sin £
C.48. Show how to solve linear equations of order higher than two by finding the general solutions of
(a) y"'  6y" + 1W ~ 62/ = 36a;, (6) 2/ (iv) + W + y = x*.
Ans. (a) 2/ = W + C2 e2x + c 3 e3x  6*  11
(6) y = d cos * + c 2 sin x + x{c 3 cos x + c 4 sin a;) + a; 2 — 4
Appendix D
Index of Special Symbols
and Notations
The following list shows special symbols and notations used in this book together with
the number of the page on which they first appear. All bold faced letters denote vectors.
Cases where a symbol has more than one meaning will be clear from the context.
Symbols
a length of semimajor axis of ellipse or hyperbola, 38, 118, 119
a n Fourier cosine coefficients, 196
a acceleration, 7
a P ip acceleration of particle P 2 relative to particle P v 7
A area, 122
A vector potential of electromagnetic field, 309
A areal velocity, 122
cA amplitude of steady state oscillation, 90
cA max maximum amplitude of steady state oscillation, 90
6 length of semiminor axis of ellipse or hyperbola, 38, 118, 119
b n Fourier sine coefficients, 196
B magnetic intensity, 83
B unit binormal, 7
c speed of light, 54
C curve, 6
D F ,D M time derivative operator in fixed and moving systems, 144
e lt e 2 , e 3 unit vectors, 72
E total energy, 36
E electric intensity, 84
f force due to friction, 65
f vX internal force on particle v due to particle X, 173
/ frequency, 89
f a frequencies, 316
F force, 33
F 12 force of particle 1 on particle 2, 33
F av average force, 60
F D damping force, 87
F v impulsive forces, 285
F„ (a) , F„ (c) actual and constraint forces acting on particle v, 170
f a generalized impulse, 285
f v force (external and internal) acting on particle v of a system of particles, 168
356
APPENDIX D] INDEX OF SPECIAL SYMBOLS AND NOTATIONS 357
g acceleration due to gravity, 62
G gravitational constant, 120
Q generating function, 314
h Planck's constant, 338
H Hamiltonian, 311
Jl Hamiltonian under a canonical transformation, 314
i unit vector in direction of positive x axis, 3
/ moment of inertia, 225
I c moment of inertia about axis through center of mass, 226
I xx , I yy , I zz moments of inertia about x, y, z axes, 254
I xy , I yz , I xz products of inertia, 254
I U I 2 ,I 3 principal moments of inertia, 255
J) v impulses, 285
$ angular impulse, 228
j unit vector in direction of positive y axis, 3
J a phase integral or action variable, 316
k unit vector in direction of positive z axis, 3
K radius of gyration, 225
I length, 90
L Lagrangian, 284
£ Lagrangian under a canonical transformation, 314
m mass, 33
m rest mass, 54
M total mass of a system of particles, 166
n number of degrees of freedom, 282
n orbital quantum number, 338
N normal component of reaction force, 65
N number of particles in a system, 166
N unit normal, 7
p a generalized or conjugate momenta, 284
p momentum, 33
P period, 89
P a new generalized momenta under a canonical transformation, 314
<? power, 34
q electrical charge, 84
q a generalized coordinates, 282
Q a new generalized coordinates under a canonical transformation, 314
r spherical coordinate, 32
r position vector or radius vector, 4
f position vector of center of mass, 166
r x unit vector in radial direction, 25
r{, position vector of particle v relative to center of mass, 169
R radius of curvature, 8
R range, 75
R max maximum range, 75
R resisting force, 64
R resultant of forces, 47
% rigid body, 228
% Routh's function or Routhian, 337
s arc length, 7
s spin angular velocity, 269
S generating function, 316
gj" generating function depending on old position coordinates and new momenta, 314
358 INDEX OF SPECIAL SYMBOLS AND NOTATIONS [APPENDIX D
t time, 6
T kinetic energy, 35
T tension, 74
T unit tangent vector, 7
T generating function depending on old and new position coordinates, 323
V generating function depending on new coordinates and old momenta, 334
v lim limiting speed, 70
v raax) v min maximum and minimum orbital speeds, 143
v velocity, 7
v velocity of center of mass, 167
Vp 2 /P i velocity of particle P 2 relative to particle P lt 7
\' v velocity of particle v relative to center of mass, 169
v i2> v i2 relative velocities of particles along common normal before and after impact, 194
V potential or potential energy, 35
"U generating function depending on old and new momenta, 334
w a angle variables, 316
W work, 34
W weight, 62
y c complementary solution, 347
y p particular solution, 347
V transverse displacement of vibrating string, 195
z cylindrical coordinate, 32
Z atomic number, 338
Greek Symbols
a angle made by vector with positive x direction, 24
a index of summation, 282
a angular acceleration, 29
/? angle made by vector with positive y direction, 24
/? damping constant, 88
f3 ratio of speed of particle to speed of light, i.e. vie, 54
7 angle made by vector with positive z direction, 24
8 logarithmic decrement, 89
8 variation symbol, 313
S a p Kronecker delta, 336
A determinant, 336
e coefficient of restitution, 195
£ eccentricity, 118
e cylindrical coordinate, 32
e Euler angle, 257
e polar coordinate, 25
6 spherical coordinate, 32
9 t unit vector perpendicular to radial direction, 25
k curvature, 8
k spring constant, 86
X colatitude, 152
^v ^2> ^3> • • • Lagrange multipliers, 280, 284
A v A 2 , A 3 components of torque along principal axes, 256
A torque or moment, 36
A c torque or moment about center of mass, 229
APPENDIX D] INDEX OF SPECIAL SYMBOLS AND NOTATIONS 359
H coefficient of friction, 65
fi reduced mass, 182
v index of summation, 166
p cylindrical coordinate, 32
p density in phase space, 318
a density, 114
a torsion, 31
t radius of torsion, 31
t time, 81
t volume, 166
<f> Euler angle, 257
<p phase angle or epoch, 88
spherical coordinate, 32
* scalar potential, 309
4> a generalized force, 283
a angular speed, 8
« lt <o 2 , « 3 components of angular velocity along principal axes, 256
u angular velocity, 144
a x , fi,,, a z components of angular momentum along x, y, z axes, 254
a lt « 2 » n 3 components of angular momentum along principal axes, 255
Q angular momentum, 37
Notations
A magnitude of A, 4
AB magnitude of distance from A to B, 11
A • B dot or scalar product of A and B, 4
A X B cross or vector product of A and B, 5
A • (B X C) scalar triple product, 5
A X (B X C) vector triple product, 5
A(w) vector function of u, 6
A(x,y,z) vector function of x,y,z, 8
0(w) scalar function of u, 6
<p(x, y, z) scalar function of x, y, z, 8
A, A time derivatives of A, i.e. dA/dt, d 2 A/dt 2 , 8
A(tt) du indefinite integral of A(m), 6
A(m) du definite integral of A(w), 6
C
I integral along curve C, 9
<s> integral around a closed path, 9
V del operator, 8
V<£ = grad <f> gradient of <p, 8
V • A = div A divergence of A, 8
V X A = curl A curl of A, 9
f(r) magnitude of central force, 116
[F, G] Poisson bracket of F and G, 331
/
INDEX
Absolute motion, 34, 60
Acceleration, 1, 7, 1720
along a space curve, 8, 20
angular, 8, 145, 148
apparent, 149
centrifugal, 145
centripetal, 8, 20, 21, 150
Coriolis, 145, 150
due to gravity, 62
in cylindrical coordinates, 32
in moving coordinate systems,
145, 146, 148, 149
in polar coordinates, 26
in spherical coordinates, 32
instantaneous, 7
linear, 145
normal, 7, 8, 19, 20
relative, 7, 18, 19
tangential, 7, 8, 19, 20
true, 149
uniform, 62, 65, 66
Action and reaction, 33
Action integral, 313
Action variables, 316, 328
Actual force, 170
Air resistance, 63, 6972
Amplitude, 86, 87
modulation, 102
of damped oscillatory motion, 88
of steadystate oscillation, 90
Angle variables, 316, 328, 329
Angular acceleration, 8, 145, 148
Angular impulse, 170, 228, 237
Angular momentum, 37, 4547
about principal axes, 255
conservation of, 37, 4547, 168, 228, 237
of a rigid body, 227, 236, 254, 259, 260
of a system of particles, 168, 169, 176, 179
of the earth about its axis, 150
principle of, 227, 229, 236, 238
relationship to torque, 37, 46,
168, 169, 176
Angular speed, 8 (see also Angular velocity)
Angular velocity, 144, 148
in terms of Euler angles, 258
of rigid body, 253
Anharmonic oscillator, 115
Aphelion, 120
Apogee, 120
Approximations, method of successive, 154, 159
Apsides, 143
Arbitrary constants, 344, 348
Arc length, 7
Areal velocity, 122, 123
Area, of parallelogram, 15
of triangle, 31
Areas, law of, 116, 123
Associative law, 3, 10
for rotations, 245
Astronomical data, 342, 343
Astronomy, definitions in, 119, 120
Asymptotes, 119
Attraction, 120, 121, 129133, 136
Atwood's machine, 76, 305
Axioms, 1
Binomial theorem, 106
Binormal, 7, 8
Bob, of simple pendulum, 90
Body axes, 257
Body centrode or locus, 229, 240, 241
Body cone, 257, 266
Bohr's quantum theory, 338
Boundary value problems, 195, 345
Bound vectors, 9, 10
Brachistochrone problem, 322
Calculus of variations, 313, 320333
connection with Hamilton's principle, 321
Canonical coordinates, 314
Canonical equations, Hamilton's, 311
Canonical transformations, 314, 323325
condition for, 314
Celestial mechanics, 311
Center, of ellipse, 118
of force, 116
of gravity, 167
of hyperbola, 119
Center of mass, 166, 172175, 183186
motion of, 167
motion of system relative to, 169, 178, 179
Central field, 116 (see also Central force)
equations of motion for particle in,
116, 122, 123
potential energy of particle in,
117, 123125
Central force, 116, 121, 122, 168, 318
(see also Central field)
determination of, from orbit, 118, 125127
Centrifugal acceleration, 145
force, 146
Centripetal acceleration, 8, 20, 21, 150
force, 146, 150
Centrode, space and body, 229, 240, 241
361
362
INDEX
Centroid {see Center of mass)
Cgs system, 33, 62, 339, 340
Chain, hanging, 186, 187
sliding, 83
Characteristic determinant, 198
frequencies, 198
Chasle's theorem, 224, 275
Circular motion, 8, 20, 21, 95, 96
Clock, 2
Coefficient of friction, 65
Colatitude, 152
Collinear vectors, 23
Collisions of particles, 194, 195, 200202
Comet, 121
Commutative law, for dot and cross
products, 3, 5, 10, 13, 14
for Poisson brackets, 331
for rotations, 230, 231
Complementary equation and solution, 347
Components, of a vector, 4
Compound pendulum, 228, 237, 238, 279, 291
Compression time, 194
Conical pendulum, 157
Conic section, 118, 127, 128
Conjugate momentum, 284, 288
Conservation, of angular momentum,
37, 4547, 168, 228, 237
of energy, 36, 4345, 124,
169, 227, 229, 236, 240
of momentum, 37, 167, 173
Conservative force fields, 35, 4345, 283,
286, 287
condition for, 35, 50, 51
Constant of the motion, 312
Constrained motion, 64, 65, 72, 73
Constraint force, 170
Constraints, 170, 180
holonomic and nonholonomic, 170, 180,
283, 284, 286, 287
reaction due to, 64, 65
Contact transformations (see Canonical
transformations)
Continuous functions, piecewise, 197
Continuous systems of particles, 165, 195
Conversion factors, table of, 341
Coordinate systems, 3, 4
inertial, 34, 39
moving, 144164
noninertial, 144
Coplanar vectors, condition for, 16
Coriolis acceleration, 145, 150
force, 146
Cosines, direction, 24
law of, 27
Couples, 226, 227, 235
Critically damped motion, 88, 96, 98, 99
Cross products, 5, 1315
determinant expression for, 14, 15
distributive law, 5, 14
failure of commutative law for, 5, 13, 14
Curl, 8, 9, 21, 22
of gradient, 9, 29
Curvature, 8, 20
radius of, 8, 20
Cycle, 87
Cyclic coordinates, 312
Cycloid, 84, 85, 106, 107, 302, 322, 323
Cycloidal pendulum, 112, 303
Cyclones, 163
Cylinder, vibrations of, 104, 105
Cylindrical coordinates, 32
acceleration in, 32
gradient in, 61
Lagrange's equations in, 291, 292
velocity and acceleration in, 32
D'Alembert's principle, 171, 182, 229
Damped harmonic oscillator, 87, 88, 9699
Damped oscillatory motion, 88, 98
Damping coefficient, 88
Damping forces, 64, 87
Deceleration, 29
Decrement, logarithmic, 89, 97, 98
Definite integrals, 6
Definitions, 1
Deformable bodies, 165
Degrees of freedom, 165, 172, 224, 225, 253, 282
of a rigid body, 172, 253, 259
Del, 8
Density, 165
in phase space, 312
Derivatives, in moving coordinate systems,
144, 145, 147, 148
notation for time, 8
of vectors, 6, 16, 17
Determinant, characteristic, 198
Jacobian, 337
secular, 198, 215
Dextral system, 4
Diagonal, main or principal, 254
Difference equation, 216
Difference of vectors, 3
Differential and derivative operators, 8, 144, 148
Differential equations, 344355
partial, 195, 344
Differential, exact, 51, 52
Dimensions, 2, 339, 340
Direct collision, 194
Direction, 2
cosines, 24
Directrix, 118
Dirichlet conditions, 197, 206, 207
Discontinuities, 197, 204, 207
Discrete system of particles, 165
Displacement, 2, 224
true, 170
virtual, 170
Dissipative forces, 64
Distance, 2
between two points, 11
Distributive law, 3
for Poisson brackets, 332
Divergence, 8, 21, 22
of curl, 9, 29
Dot products, 4, 5, 12, 13
INDEX
363
Dot products (cont.)
commutative law for, 5
distributive law for, 5, 12
Double pendulum, 285, 286, 299301
Drumhead, vibrating, 195
Dumbbell, 278
Dynamics, 1
Dyne, 33
Earth, flat, 63
motion of particle relative to, 145
rotation of, 150, 257, 265
Eccentricity, 118
Einstein's laws of relativity, 34, 61
Elastic bodies, 165
Elastic collisions, perfectly, 195, 201
Elastic constant, 86
Elasticity, 194
modulus of, 86
Elastic string, vibrations of
(see Vibrating string)
Electrical charge, 83, 84
Electromagnetic field, 84, 309
Hamiltonian for particle in, 338
Lagrangian for particle in, 309
Ellipse, 38, 104, 118, 119, 121, 127, 128
Ellipsoid of inertia, 255, 256, 263, 264
Elliptic functions, 106, 272, 279
integrals, 106, 108
Energy, conservation of, 36, 4345, 124, 169,
227, 229, 236, 240
kinetic, (see Kinetic energy)
of simple harmonic oscillator, 87, 99
potential (see Potential energy)
total, 36
English systems, 63, 339
Epoch, 87, 88, 93
Equality of vectors, 2
Equilibrant, 47
Equilibrium, 37, 171
in a uniform gravitational field, 65, 66, 74, 75
of a particle, 37, 38, 47, 48
of a rigid body, 229, 241, 242
of a system of particles, 170, 180, 181
position, 86
stable, 38, 48, 49, 60, 141, 171, 230
Escape speed or velocity, 134
Euclidean geometry, 1, 2
Euler angles, 253, 257, 267, 268, 301
angular velocity in terms of, 258
Euler's equations of motion, 256, 264
from Lagrange's equations, 302
Euler's or Lagrange's equations, 313, 320
Euler's theorem, 224
on homogeneous functions, 305, 306, 317
Even extension of a function, 208
Even functions, 196
Event, 2
E volute, 112
Exact differential, 51, 52
Exact differential equation, 345, 350
Extremal, 313
Extremum or extreme value, 313
Field, scalar or vector, 8
Flat earth, 63 (see also Earth)
Focus, 118
Force, 33
axiomatic definition of, 33, 49
centrifugal and centripetal, 145, 146, 150
constraint, 170
damping, 64, 87
generalized, 283
units of, 33, 339, 340
Forced vibrations, 89, 99102
resonance and, 90, 100, 101
Force fields, conservative (see Conservative
force fields)
nonconservative, 37, 47
uniform, 62, 65, 66
Foucault pendulum, 146, 154156
Fourier coefficients, 196, 206
Fourier series, 195197, 203208
convergence of, 197
half range, 197, 207, 208
Fourier coefficients for, 196, 206
solution of vibrating string by
(see Vibrating string)
Fps system, 33, 62, 339, 340
Frames of reference, Newtonian, 33, 34
Freely falling bodies, 63, 67
Free vectors, 9, 10
FrenetSerret formulas, 31
Frequencies, characteristic, 198
Frequency, fundamental, 211
natural, 89, 98
obtained by Hamiltonian methods, 316, 329
of precession, 257, 265, 270, 273, 274
of resonance, 90
of simple harmonic motion, 86, 87
Friction, 65
coefficient of, 65
motion involving, 73
Fss system, 63, 339, 340
Function, scalar and vector, 8
Fundamental frequency, 211
Generalized coordinates, 282, 285, 286
forces, 283
impulse, 285
momenta, 284, 288
velocities, 283
General solution of differential equation,
344, 348
Generating functions, 314, 315, 323325
Geometry, Euclidean, 1, 2
Gimbal, of a gyroscope, 258
Gradient, 8, 21, 22
curl of, 9, 29
in cylindrical and spherical coordinates, 61
Gram, 33
Gravitation, universal law of (see Universal
law of gravitation)
Gravitational constant, universal, 120
Gravitational potential, 120, 121, 133, 143
Gravitational system of units, 63, 339
364
INDEX
Gravity, 62
center of, 167
vibrating string under, 214, 215
Gyration, radius of, 225
Gyrocompass, 278
Gyroscopes, 258, 268273
Half range Fourier sine and cosine series
197, 207, 208
Hamiltonian, 311, 317, 318, 330
for conservative systems, 311
for particle in electromagnetic field, 338
HamiltonJacobi equation, 315, 325327, 330, 331
for Kepler's problem, 326, 327
for one dimensional harmonic oscillator,
325, 326
solution of, 315, 316
Hamilton's equations, 311, 317, 318
Hamilton's principle, 313, 320323
Harmonic oscillator, damped, 87, 88, 9699
simple, 8690, 92102
two and three dimensional, 91, 103, 104
Herpolhode, 257, 266
Holonomic, 170, 180, 283, 284, 286, 287
Homogeneous equation, 346, 351
Homogeneous functions, 305
Euler's theorem on, 305, 306, 317
Hooke's law, 86
Hyperbola, 104, 119, 121, 127, 128
Hyperbolic functions, 54
Ignorable coordinates, 312, 315
Impact, 194
Impulse, 36, 4547, 169, 170, 180
angular, 170, 228, 237
generalized, 285
relation to momentum, 36
Impulsive forces, 285, 295298
Inclined plane, 64, 65, 72
motion of particle down, 72, 73
motion of sphere down, 239, 240
projectile motion on, 75, 76, 81
Incompressible fluid, 313
Indefinite integrals of vectors, 6
Independence of path, 9
condition for, 50, 51
Inelastic collisions, perfectly, 195, 201
Inertial frames of reference, 33, 34, 39
classical principle of relativity for, 39
Inertial system, 34, 39
Initial point, of a vector, 2
Instability, 38
Instantaneous, acceleration, 7
axis of rotation, 224, 225, 229
center of rotation, 225, 229, 240, 241
power, 34 (see also Power)
velocity, 7
Integral equations, 154
Integrals of vectors, 6, 16, 17
line (see Line integrals)
Integrating factor, 345, 351
Internal forces, 177, 178
Invariable line and plane, 256, 257, 266
Invariant, 34
Isolation, of a system, 64
Iteration, method of, 154, 159
Jacobian determinant, 337
Kepler's laws, 120, 128, 129, 223
deduction of from Newton's universal
law of gravitation, 125, 126, 129
Kilogram, 33
weight, 63
Kinematics, 1
Kinetic energy, 34, 35, 4143
about principal axes, 255
in terms of Euler angles, 258, 268
in terms of generalized velocities, 283, 287, 288
of a rigid body, 227, 236, 259, 260
of a system of particles, 168, 169, 179, 182
of rotation, 229
of translation, 229
relationship to work, 35, 41, 169
relativistic, 55
Kronecker delta, 336
Lagrange multipliers, 280, 284, 292, 295
Lagrange's equations, 282310, 320
and calculus of variations (see Calculus
of variations)
for conservative systems, 284, 288292
for nonconservative systems, 284
for nonholonomic systems, 284, 285,
292295, 303
with impulsive forces, 285, 295298
Lagrangian function, 284, 311
for particle in electromagnetic field, 309
Latitude, 152
Laws, 1
Lemniscate, 138
Length, 2
Light, speed of, 34, 54
Limiting speed or velocity, 70, 72
Line, 1
of action of a vector, 10
Linear equations, 345347, 349, 350, 352
Linear harmonic oscillator, 86 (see also
Harmonic oscillator)
Linear impulse (see Impulse)
Linear momentum (see Momentum)
Line integrals, 9, 22, 23
evaluation of, 22, 23
independence of path of, 9, 23
Liouville's theorem, 312
proof of, 318320
Lissajous curves or figures, 91
Logarithmic decrement, 89, 97, 98
Lorentz force, 84
Magnetic field, 83
Main diagonal, of moment of inertia
matrix, 254
Major axis, of ellipse, 118
of hyperbola, 139
INDEX
Mass, 2, 33
axiomatic definition of, 49
center of (see Center of mass)
changing, 194
of the earth, 129
reduced, 182, 231
rest, 54, 61
units of, 33
Mathematical models, 1
Matrix, moment of inertia, 254
Matter, 1, 2
Mechanics, 1
relativistic, 34
Membrane, vibrating, 195
Meteorite, 121
Metric system, 339
Minor axis, of ellipse, 119
of hyperbola, 139
Mks system, 33, 62, 339, 340
Mode of vibration, normal, 194
Models, mathematical, 1
Modulation, amplitude, 102
Modulus of elasticity, 86
Moment, of couple, 226
of force, 36
of momentum, 37 (see also Angular momentum)
Momental ellipsoid, 256
Moments of inertia, 225, 231233, 254, 259,
260, 263, 264
matrix, 254
principal (see Principal moments of inertia)
special, 226
theorems on, 225, 233235
Momentum, 33, 167
angular (see Angular momentum)
conjugate, 284, 288
conservation of, 37, 167, 173
generalized, 284, 288
moment of, 37 (see also Angular momentum)
of a system of particles, 167, 169, 172, 173
principle of, 238
Momentum coordinates, 312
Moon, 119, 342
Multiplyperiodic vibrations, 194
Musical note, 211
Natural frequency and period, 89, 98
Newton, 33
Newtonian frames of reference, 34
Newton's collision rule, 194, 202
laws of motion, 3341
universal law of gravitation (see
Universal law of gravitation)
Nodes, line of, 257
Nonholonomic, 170, 180, 283, 284, 286, 287
Noninertial systems, 144, 145
Normal frequencies, 194, 198
for a double pendulum, 300, 301, 308
for a vibrating string, 210, 211
for a vibrating system of particles, 215, 216
Normal modes of vibration, 194, 198, 199
for a vibrating string, 210, 211
Normal, principal, 7, 8, 20
Normal (cont.)
to a surface, 24
Null vector, 3
Nutation, 270, 272
Oblique collisions, 194
Odd extension, of a function, 207
Odd functions, 196
Operators, derivative, 144
Optics, 335
Orbit, 116, 120
determination of from central force,
117, 118, 125127
weightlessness in, 135, 136
Order of a differential equation, 344
Oscillations, forced, 89 (see also
Forced vibrations)
Oscillator, anharmonic, 115
harmonic (see Harmonic oscillator)
Overdamped motion, 88, 98, 99
Overtones, 211
Pappus, theorems of, 193
Parabola, 63, 104, 119, 121, 127, 128
as curve of motion of projectile, 68
Paraboloid of revolution, 107, 108
Parachutist, motion of, 69, 70
Parallel axis theorem, 226, 233, 234
Parallelepiped, volume of, 5, 15, 16
Parallelogram, area of, 5, 15
Parallelogram law, 2
Partial differential equation, 195, 344
of vibrating string (see Vibrating string)
Particles, 2
equilibrium of, 37, 38
systems of, 165193
vibrations of, 194, 197199
Particular solutions, 344, 345, 347, 348
Path, independence of, 9, 50, 51
Pendulum, bob, 90
compound, 228, 237, 238, 279, 291
conical, 157
cycloidal, 112, 303
double, 285, 286, 299301
Foucault, 146, 154156
seconds, 110
simple (see Simple pendulum)
Perigee, 120
Perihelion, 120
Period, 53
natural, 89
of damped motion, 89
of harmonic oscillator, 87
of motion in a magnetic field, 83
of simple harmonic motion, 86, 87
of simple pendulum, 91, 105, 106
orbital, 135, 136
sidereal, 120
Perpendicular axes theorem, 226, 234, 235
Phase, angle, 87, 88, 93
integrals, 316, 328, 329
out of, 93
space, 312, 318320
366
INDEX
Piano string, vibrations of, 195
(see also Vibrating string)
Piecewise continuous functions, 197
Planck's constant, 338
Planets, 119, 343
Poinsot's construction, 257
Point, 1, 2
Poisson bracket, 331, 332
Polar coordinates, 25, 26
gradient in, 54
velocity and acceleration in, 26
Polhode, 257, 266
Position, 2
coordinates, 312
vector, 4
Potential, 35 (see also Potential energy)
relation to stability, 38
scalar, 35, 309
vector, 309
Potential energy, 35, 36, 4345
(see also Potential)
in a central force field, 117, 123125
in a uniform force field, 64, 69
of a system of particles, 169, 176178
principle of minimum, 230
relation of to work, 35, 44
Pound, 33
weight, 63
Poundal, 33
Power, 34, 4143, 227, 237
relation to work, 42
Precession, 156, 256, 270, 272
frequency of, 257, 265, 270, 273, 274
Principal axes of inertia, 255, 260263
Principal diagonal, 254
Principal moments of inertia, 255, 260263
method of Lagrange multipliers for, 280
Principal normal, unit, 7, 8, 20
Products of inertia, 254, 259, 260
Products of vectors, by a scalar, 3
cross (see Cross products)
dot (see Dot products)
Projectiles, 62, 63
maximum height of, 68
motion of, 68, 69, 71, 72
on an inclined plane, 75, 76, 81
range of (see Range of projectile)
time of flight, 68
Pulley, 76, 289, 290
Quantum mechanics, 311
Quantum number, orbital, 338
Quantum theory, 338
Radius, of curvature, 8, 20
of gyration, 225
of torsion, 31
Radius vector, 4
Range of projectile, 68
maximum, 69
on inclined plane, 75, 76
on rotating earth, 164
Reaction, 33
Reaction (cont.)
due to constraints, 64, 65
Rectangular coordinate systems, 3, 4
right handed, 4
Reduced mass, 182, 231
Reference level, 64
Relative acceleration, 7, 18, 19
velocity, 7
Relativistic mechanics, 34
Relativity, classical principle of, 34, 39
Einstein's laws of, 34, 61
theory of 54, 55, 61, 143, 337
Representative point, 312
Resistance, air, 63, 6972
Resisting forces, 64
Resisting medium, motion in, 64, 6972
Resonance, 90, 100, 101
Restitution, coefficient of, 195
Restitution time, 194
Rest mass, 54, 61
Restoring force, 86
Resultant of vectors, 2
Rheonomic, 283, 286, 287
Right handed system, 4
Rigid bodies, 165, 170, 224, 230, 231, 236
equilibrium of, 229, 241, 242
force free motion of, 256, 257, 265
general motion of, 224, 253, 259
motion of, about a fixed axis, 236
plane motion of, 224252
symmetric, 257
Rockets, 173, 194, 199, 200
motion of, 199, 200
Rotating coordinate systems, 144, 147, 148
Rotation, 224, 253
associative and commutative laws for, 230,
231, 245
finite, 230, 231
of the earth, 150, 257, 265
pure, 253
Routh's function or Routhian, 337
Satellites, 119
Scalar function, 8
Scalar potential, 35
for electromagnetic field, 309
Scalar product (see Dot product)
Scalar triple product, 5
Scalars, 2
Scleronomic, 283, 286, 287
Seconds pendulum, 110
Secular determinant, 198, 215
Semimajor and minor axes, 118, 119, 129
Separation of variables, 210, 316, 345, 347, 348
Sidereal period, 120
Simple closed curve, 9
Simple harmonic motion, 86
(see also Simple harmonic oscillator)
amplitude, period and frequency of, 86, 87
Simple harmonic oscillator, 8690, 92102
amplitude, period and frequency of, 86, 87
damped, 87, 88
energy of, 87, 99
INDEX
367
Simple harmonic oscillator (cont.)
forced vibrations of, 89, 99102
resonance and, 90, 100, 101
Lagrange's equations for, 306
Simple pendulum, 86, 87, 90, 91, 102, 103
length of equivalent, 228
Sines, law of, 27
Sliding vector, 9, 10
Slug, 63
Solar system, 119
Solution of differential equation, 344
Space, 1, 2
Space axes, 257
Space centrode or locus, 229, 240, 241
Space cone, 257, 266
Special relativity (see Relativity)
Speed, 7 (see also Velocity)
angular, 8
escape, 134
of light, 34
orbital, 135, 136
Sphere, particle sliding down, 76, 77, 82
sphere rolling down, 244, 303, 304
Spherical coordinates, 32
gradient in, 61
Lagrange's equations in, 306
velocity and acceleration in, 32
Spin, 270, 272
Spring constant, 86
Spring, vibrations of, 86, 9395
Stability of equilibrium, 38, 48, 49, 60, 141,
171, 230
Stable point, 38
Star, 119
Statics, 1, 37, 38, 47
in a uniform gravitational field, 65, 66, 74, 75
of a particle, 37, 38, 47, 48
of a rigid body, 229, 241, 242
of a system of particles, 170, 180, 181
Statistical mechanics, 311
Steadystate solution, 89
Stiffness factor, 86
Sum of vectors, 2
obtained graphically and analytically, 12, 48
Sun, 119, 342
Superposition principle, 199
Surface, normal to, 24
Symmetric matrix or tensor, 254
Systems of particles, 165193
Tangent vector, unit, 7, 8, 19
Tautochrone problem, 113
Tension, 74, 76
Tensor, moment of inertia, 254
Terminal point, of a vector, 2
Theorems, 1
Time. 1, 2
principle of least, 335
Top, 258, 268273, 274
Lagrange's equations for motion of, 301, 302
motion of, 258, 268273
sleeping, 274
Top (cont.)
steady precession of, 270
Torque, 36, 4547, 168, 176
of a couple, 226
relation to angular momentum, 37, 46, 168,
169, 176
Torsion, 31
constant of, 308
radius of, 31
Transformation equations, 282, 285, 286
canonical, 314, 323325
Transient solution, 89
Translation, 224, 253
Transverse vibration of a string
(see Vibrating string)
Triple products, 5, 15, 16
scalar, 5
vector, 5
Two and three body problems, 121, 223
Underdamped motion, 88, 98
Undetermined coefficients, method of, 347,
352, 353
Uniform acceleration, 62, 65, 66
force field, 62, 65, 66
Uniformly accelerated motion, 62, 65, 66
Units, 2, 339341
Unit vectors, 3
rectangular, 3, 4
Universal law of gravitation, 120, 128, 129
deduction from Kepler's laws, 128, 129
Unstable equilibrium, 171, 230
Variation of an integral, 321
Variation of parameters, method of, 347, 353, 354
Variation symbol, 313, 337
Variations, calculus of
(see Calculus of variations)
Vector algebra, laws of, 3, 1012
Vector field, 8
Vector function, 8
Vector potential, 309
Vector product (see Cross product)
Vector triple product, 5
Vectors, 1, 2
algebra of, 2, 3
bound, 9, 10
components of, 4
definition of, 2
free, 9, 10
magnitude of, 11, 13
sliding, 9, 10
Velocity, 1, 6, 7, 1719
angular (see Angular velocity)
apparent, 148
areal, 122, 123
escape, 134
generalized, 283
in cylindrical coordinates, 32
in moving cordinate systems, 145, 148, 149
in polar coordinates, 26
in spherical coordinates, 32
368
INDEX
Velocity (cont.)
instantaneous, 7
limiting, 70, 72
of a rigid body, 253, 259
relative, 7
true, 148
Vertices, of ellipse, 118
of hyperbola, 119
Vibrating string, 195, 202, 203, 209212
considered as a system of particles, 215217
under gravity, 214, 215
Vibrating systems of particles, 194, 197199
Vibrations, of a cylinder, 104, 105
forced (see Forced vibrations)
Violin string, vibrations of, 195
(see also Vibrating string)
Virtual displacements, 170
Virtual work, Lagrange's equations and, 292
Virtual work, principle of, 170, 229
Weight, 62
apparent, 162
Weightlessness, 135, 136
Work, 34, 4143, 168, 169, 176178, 237
generalized forces and, 283, 287, 288
in rotation of a rigid body, 227
relationship of to kinetic energy, 168, 169,
176, 177
relationship of to potential energy, 44
virtual (see Virtual work)
x direction, 4
y direction, 4
z direction, 4
Zero vector, 2